8 Inference for One Sample

Statistical Inference

One Proportion

One Mean

Paired Means

t-Distributions

Objectives

Upon completion of this lesson, you should be able to:

Identify situations in which the z- or t-distributions may be used to approximate a sampling distribution.
Construct a confidence interval to estimate a population proportion, mean, or difference in paired means by hand given summary data.
Construct a confidence interval to estimate a population proportion, mean, or difference in paired means using Minitab given summary or raw data.
Determine the necessary minimum sample size to construct a confidence interval for a single proportion or single mean with a given level of confidence and margin of error.
Conduct a hypothesis test using the appropriate common distribution for a single proportion, single mean, and paired means by hand given summary data.
Conduct a hypothesis test using the appropriate common distribution for a single proportion, single mean, and paired means using Minitab given summary or raw data.

This lesson corresponds to Sections 6.1, 6.2, and 6.5 in the Lock⁵ textbook.

The general form of confidence intervals and test statistics will be the same for all the procedures covered in this lesson:

General Form of a Confidence Interval

\[ \text{sample statistic}\pm(\text{multiplier})\ (\text{standard error}) \]

General Form of a Test Statistic

\[ \text{test statistic}=\dfrac{\text{sample statistic}-\text{null parameter}}{\text{standard error}} \]

We will be using a five-step hypothesis testing procedure:

Check assumptions and write hypotheses. The assumptions will vary depending on the test. The null and alternative hypotheses will also be written in terms of population parameters; the null hypothesis will always contain the equality (i.e., =).
Calculate the test statistic. This will vary depending on the test, but it will typically be the difference observed between the sample and population divided by a standard error. In this lesson, we will see z- and t-test statistics. Minitab will compute the test statistic.
Determine the p-value. This can be found using Minitab.
Make a decision. If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis.
State a “real world” conclusion. Based on your decision in step 4, write a conclusion in terms of the original research question.

Some steps may vary depending on the test.

We will be relying heavily on Minitab in this lesson and all the following lessons. Note that help is available in Minitab.

8.1 One Sample Proportion

One sample proportion tests and confidence intervals are covered in Section 6.1 of the Lock⁵ textbook.

In the last lesson you were introduced to the general concept of the Central Limit Theorem. The Central Limit Theorem states that if the sample size is sufficiently large then the sampling distribution will be approximately normally distributed for many frequently tested statistics, such as those that we have been working with in this course. When discussion proportions, we sometimes refer to this as the Rule of Sample Proportions. According to the Rule of Sample Proportions, if $np\geq 10$ and $n(1-p) \geq 10$ then the sampling distributing will be approximately normal. When constructing a confidence interval $p$ is not known but may be approximated using $\widehat p$. When conducting a hypothesis test, we check this assumption using the hypothesized proportion (i.e., the proportion in the null hypothesis).

If assumptions are met, the sampling distribution will have a standard error equal to $\sqrt{\frac{p(1-p)}{n}}$.

This method of constructing a sampling distribution is known as the normal approximation method.

If the assumptions for the normal approximation method are not met (i.e., if $np$ or $n(1-p)$ is not at least 10), then the sampling distribution may be approximated using a binomial distribution. This is known as the exact method. This course does not cover the exact method in detail, but you will see how these tests may be performed using Minitab.

8.1.1 Confidence Intervals

In the following section you will see how a confidence interval for a population proportion can be constructed by hand using the normal approximation method. Using Minitab, you will learn how to construct a confidence interval for a proportion using the normal approximation method or the exact method. When given the option, it is recommended that you use Minitab as opposed to performing calculations by hand.

Normal Approximation Formulas

For the following procedures, the assumption is that both $np \geq 10$ and $n(1-p) \geq 10$. When we’re constructing confidence intervals $p$ is typically unknown, in which case we use $\widehat{p}$ as an estimate of $p$.

Note that $n \widehat p$ is the number of successes in the sample and $n(1- \widehat p)$ is the number of failures in the sample.

This means that our sample needs to have at least 10 “successes” and at least 10 “failures” in order to construct a confidence interval using the normal approximation method.

Below is the general form of a confidence interval.

General Form of Confidence Interval

\[ \text{sample statistic}\pm\underbrace{(\text{multiplier})\ (\text{standard error})}_{\textbf{margin of error}} \]

The sample statistic here is the sample proportion, $\widehat p$. When using the normal approximation method the multiplier is taken from the standard normal distribution (i.e., z-distribution). And, the standard error is computed using $\widehat p$ as an estimate of $p$:

\[ \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} \]

This leaves us with the following formula to construct a confidence interval for a population proportion:

Confidence Interval of $p$: Normal Approximation Method

\[ \underbrace{\widehat{p}}_{\text{sample statistic}} \pm \overbrace{z^{*}}^{\text{multiplier}} \underbrace{\left (\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)}_{\text{standard error}} \]

Finding the $z^{*}$ Multiplier

The value of the $z^*$ multiplier depends on the level of confidence. The multiplier for the confidence interval for a population proportion can be found using the standard normal distribution [i.e., z-distribution, N(0,1)]. The most commonly used level of confidence is 95%. As shown on the probability distribution plot below, the multiplier associated with a 95% confidence interval is 1.960, often rounded to 2 (recall the empirical rule and 95% rule).

Standard normal distribution showing the z multipliers for a 95% confidence interval. — Fig 8.1

Below is a table of frequently used $z^*$ multipliers.

Confidence Level	$z^*$ Multiplier
90%	1.645
95%	1.960, often rounded to 2
98%	2.327
99%	2.576

The value of the multiplier increases as the confidence level increases. This leads to wider intervals for higher confidence levels. We are more confident of catching the population value when we use a wider interval.

Example 8.1 (PA Residency)

Video 8.1: Confidence Interval for a Proportion, PA Residency Example

Example 8.2 (Dog Ownership) In Spring 2016, a sample of 522 World Campus students were surveyed and asked if they own a dog. Of the 522 students in the sample, 273 said that they did have a dog. Construct a 95% confidence interval for the proportion of all World Campus students who have a dog.

Video 8.2: 95% Confidence Interval for Proportion of Dog Owners

Example 8.3 (Books)

Video 8.3: STAT 200 95% Confidence Interval for P Example: Paper Copy of Textbook

Example 8.4 (Retirement) In a representative sample of 1168 American adults, 747 said they were not financially prepared for retirement. Let’s construct a 95% confidence interval to estimate the proportion of all American adults who are not financially prepared for retirement.

First, we need to check our assumptions that both $n\widehat p \geq 10$ and $n(1-\widehat p) \geq 10$.

\[ \widehat{p}=\frac{747}{1168}=0.640 \]

\[ np=1168 (0.640) = 747 \text{ and } n(1-\widehat p)=1168(1-0.640)=421 \]

Both are greater than 10, so this assumption has been met. This means we can use the normal approximation method to construct this confidence interval.

Next, we can compute the standard error.

\[ \text{SE}=\sqrt{\frac{\hat{p} (1-\hat{p})}{n}}=\sqrt{\frac{0.640 (1-0.640)}{1168}}=0.014 \]

The $z^*$ multiplier for a 95% confidence interval is 1.960.

The formula for a confidence interval for a proportion is $\widehat{p}\pm z^* (\text{SE})$.

\[ 0.640\pm 1.960(0.014)=0.640\pm0.028=[0.612, 0.668] \]

We are 95% confident that between 61.2% and 66.8% of all American adults are not financially prepared for retirement.

What if we wanted a 99% confidence interval?

Let’s think about how our interval will change. The 99% confidence interval will be wider than the 95% confidence interval. In order to increase our level of confidence, we will need to expand the interval.

In terms of computing the 99% confidence interval, we will use the same point estimate $\widehat{p}$ and the same standard error. Only the multiplier will change. From the plot below, we see that the $z^*$ multiplier for a 99% confidence interval is 2.576.

Standard normal distribution showing the z multipliers for a 99% confidence interval — Fig 8.2

99% CI: $0.640\pm 2.576 (0.014)=0.0640\pm 0.036=[0.604, 0.676]$

We are 99% confident that between 60.4% and 67.6% of all American adults are not financially prepared for retirement.

Minitab: Confidence Interval for a Proportion

Construct a one-proportion confidence interval with raw data in Minitab:

Select Stat > Basic Statistics > 1 Proportion.
Select the variable containing the raw data.
Select the Options tab.
Enter the appropriate confidence level.
Set the alternative hypothesis to not equal.
Under Method, select Normal approximation.
Select OK.

Before we can construct a confidence interval for a proportion, we must first determine if we should use the exact method or the normal approximation method. Recall that if $np \geq 10$ and $n(1-p) \geq 10$, then the sampling distribution can be approximated by a normal distribution. Since we don’t have the population proportion ($p$), we will use $\widehat p$ as an estimate. Note that $n\widehat p$ is the number of successes in the sample and $n(1-\widehat p)$ is the number of failures in the sample.

If this assumption has not been met, then the sampling distribution is constructed using a binomial distribution which Minitab refers to as the “exact method.”

To check this assumption, we can construct a frequency table. You first learned how to construct a frequency table in Section 2.1.1.2 of these online notes. Here is another example:

Example 8.5 (Minitab: Frequency Tables) To create a frequency table of dog ownership in Minitab:

Open the data set: (fall2016stdata.mpx)
From the toolbar in Minitab, select Stat > Tables > Tally Individual Variables.
Insert the variable Dog in the box on the left into the Variable box on the right.
Under Display, choose Counts.
Select OK.

This should result in the following frequency table:

Dog	Count
No	252
Yes	272
N=	524
*=	1

From the frequency table above we can see that there were at least 10 “successes” and at least 10 “failures” in the sample. In this example a success is defined as answering “yes” to the question “do you own a dog?” A failure is defined as answering “no.” Because both $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$, the normal approximation method may be used. In Minitab, the exact method is the default method. If there are at least 10 successes and at least 10 failures, then you need to change the method to the normal approximation method.

Example 8.6 (Minitab: Confidence Interval for a Proportion (Normal Approximation)) To create a 95% confidence interval of dog ownership using the normal approximation method in Minitab:

Open the data set: (fall2016stdata.mpx)
In Minitab, select Stat > Basic Statistics > 1-Proportion.
In this case we have our data in the Minitab worksheet so we will use the default One or more samples each in a column.
Insert the variable Dog in the box on the left into the box on the right.
Select Options.
The default Confidence level is 95.
Change the Method to Normal approximation because the assumption of $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$ has been met.
Select OK.

This should result in the following output:

Method

Event: Dog = Yes

p: proportion where Dog = Yes

Normal approximation is used for this analysis.

Descriptive Statistics

N	Event	Sample p	95% CI for p
524	272	0.519084	(0.476304, 0.561863)

Considerations

What if the assumption is not met?

If the number of successes or the number of failures in the sample is less than 10, then the exact method should be used instead of the normal approximation method. In Minitab, this means that in step 8 above the default setting of Exact method should not be changed.

What if we have summarized data and not data in a Minitab worksheet?

If you do not have a Minitab worksheet filled with data concerning individuals, but instead have summarized data (e.g., the number of successes and the number of failures), you would not load the data set, but in step 3 you would select Summarized data. For Number of events, enter the number of successes (i.e., $n \widehat p$) and for Number of trials, enter the total sample size (i.e., $n$).

Try It!

(Lactose Intolerance) In a sample of 100 African American adults, 70 were identified as having some level of lactose intolerance. Use Minitab to compute a 95% confidence interval to estimate the proportion of all African American adults who have some level of lactose intolerance.
To create a 95% confidence interval of lactose intolerance using the normal approximation method in Minitab:
1. In Minitab, select Stat > Basic Statistics > 1-Proportion.
2. In this case we have summarized data so select Summarized data in the dropdown.
3. For number of events, add 70 and for number of trials add 100.
4. Select Options.
5. The default Confidence level is 95.
6. Change the Method to Normal approximation because the assumption of $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$ has been met.
7. Select OK, then OK.
This should result in the following output:

Method

p: event proportion

Normal approximation is used for this analysis.

Descriptive Statistics

N Event Sample p 95% CI for p

100 70 0.700000 (0.610183, 0.789817)
(Dieting) At the beginning of the Fall 2016 semester a representative sample of World Campus STAT 200 students was surveyed. The students were asked if they were currently dieting to lose weight. In the sample of 524 students, 184 said that they were dieting to lose weight. Use Minitab to construct a 95% confidence interval for the proportion of all World Campus STAT 200 students who are dieting to lose weight.
1. In Minitab, select Stat > Basic Statistics > 1-Proportion.
2. In this case we have summarized data, so select Summarized data in the dropdown.
3. For number of events, add 184 and for number of trials add 524.
4. Select Options.
5. The default Confidence level is 95.
6. Change the Method to Normal approximation because the assumption of $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$ has been met.
7. Select OK, then OK.
This should result in the following output:

Method

p: event proportion

Normal approximation is used for this analysis.

Descriptive Statistics

N Event Sample p 95% CI for p

524 184 0.351145 (0.310276, 0.392015)

Descriptive Statistics
N	Event	Sample p	95% CI for p
100	70	0.700000	(0.610183, 0.789817)

N	Event	Sample p	95% CI for p
524	184	0.351145	(0.310276, 0.392015)

8.1.2 Computing Necessary Sample Size

When we begin a study to estimate a population parameter we typically have an idea as how confident we want to be in our results and within what degree of accuracy. This means we get started with a set level of confidence and margin of error. We can use these pieces to determine a minimum sample size needed to produce these results by using algebra to solve for $n$:

Finding Sample Size for Estimating a Population Proportion

\[ n=\left ( \dfrac{z^*}{M} \right )^2 \tilde{p}(1-\tilde{p}) \]

$M$ is the margin of error
$\tilde p$ is an estimated value of the proportion

If we have no preconceived idea of the value of the population proportion, then we use $\tilde{p}=0.50$ because it is most conservative, and it will give use the largest sample size calculation.

Example 8.7 (No Estimate) We want to construct a 95% confidence interval for $p$ with a margin of error equal to 4%.

Because there is no estimate of the proportion given, we use $\tilde{p}=0.50$ for a conservative estimate.

For a 95% confidence interval, $z^*=1.960$.

\[ n=\left ( \dfrac{1.960}{0.04} \right )^2 (0.5)(1-0.5)=600.25 \]

This is the minimum sample size, therefore we should round up to 601. In order to construct a 95% confidence interval with a margin of error of 4%, we should obtain a sample of at least $n=601$.

Example 8.8 (Estimate Known) We want to construct a 95% confidence interval for $p$ with a margin of error equal to 4%. What if we knew that the population proportion was around 0.25?

The $z^*$ multiplier for a 95% confidence interval is 1.960. Now, we have an estimate to include in the formula:

\[ n=\left ( \dfrac{1.960}{0.04} \right )^2 (0.25)(1-0.25)=450.188 \]

Again, we should round up to 451. In order to construct a 95% confidence interval with a margin of error of 4%, given $\tilde{p}=.25$, we should obtain a sample of at least $n=451$.

Note that when we changed $\tilde{p}$ in the formula from .50 to .25, the necessary sample size decreased from $n=601$ to $n=451$.

8.1.3 Hypothesis Testing

A hypothesis test for a proportion is used when you are comparing one group to a known or hypothesized population proportion value. In other words, you have one sample with one categorical variable. The hypothesized value of the population proportion is symbolized by $p_0$ because this is the value in the null hypothesis ($H_0$).

If $np_0 \ge 10$ and $n(1-p_0) \ge 10$ then the distribution of sample proportions is approximately normal and can be estimated using the normal distribution. That sampling distribution will have a mean of $p_0$ and a standard deviation (i.e., standard error) of $\sqrt{\frac{p_0 (1-p_0)}{n}}$.

Recall that the standard normal distribution is also known as the z-distribution. Thus, this is known as a “single sample proportion z-test” or “one sample proportion z-test.”

If $np_0 < 10$ or $n(1-p_0) < 10$ then the distribution of sample proportions follows a binomial distribution. We will not be conducting this test by hand in this course, however you will learn how this can be conducted using Minitab using the exact method.

Normal Approximation Method Formulas

Here we will be using the five-step hypothesis testing procedure to compare the proportion in one random sample to a specified population proportion using the normal approximation method.

Check assumptions and write hypotheses.

In order to use the normal approximation method, the assumption is that both $n p_0 \geq 10$ and $n (1-p_0) \geq 10$. Recall that $p_0$ is the population proportion in the null hypothesis.

Research Question	Is the proportion different from $p_0$?	Is the proportion greater than $p_0$?	Is the proportion less than $p_0$?
Null Hypothesis, $H_{0}$	$p=p_0$	$p= p_0$	$p= p_0$
Alternative Hypothesis, $H_{a}$	$p\neq p_0$	$p> p_0$	$p< p_0$
Type of Hypothesis Test	Two-tailed, non-directional	Right-tailed, directional	Left-tailed, directional

Where $p_0$ is the hypothesized population proportion that you are comparing your sample to.

Calculate the test statistic.

When using the normal approximation method we will be using a z-test statistic. The z-test statistic tells us how far our sample proportion is from the hypothesized population proportion in standard error units. Note that this formula follows the basic structure of a test statistic that you learned in the last lesson:

\[ \text{test statistic}=\dfrac{\text{sample statistic}-\text{null parameter}}{\text{standard error}} \]

Test statistic: One Group Proportion

\[ z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}} \tag{8.1}\]

$\widehat{p}$ = sample proportion
$p_{0}$ = hypothesize population proportion
$n$ = sample size
Determine the p-value.

Given that the null hypothesis is true, the p-value is the probability that a randomly selected sample of n would have a sample proportion as different, or more different, than the one in our sample, in the direction of the alternative hypothesis. We can find the p-value by mapping the test statistic from step 2 onto the z-distribution.

Note!
Note that p-values are also symbolized by $p$. Do not confuse this with the population proportion which shares the same symbol.

We can look up the $p$-value using Minitab by constructing the sampling distribution. Because we are using the normal approximation here, we have a z-test statistic that we can map onto the z-distribution. Recall, the z-distribution is a normal distribution with a mean of 0 and standard deviation of 1. If we are conducting a one-tailed (i.e., right- or left-tailed) test, we look up the area of the sampling distribution that is beyond our test statistic. If we are conducting a two-tailed (i.e., non-directional) test there is one additional step: we need to multiply the area by two to take into account the possibility of being in the right or left tail.
Make a decision.

We can decide between the null and alternative hypotheses by examining our p-value. If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. Unless stated otherwise, assume that $\alpha=.05$.

When we reject the null hypothesis our results are said to be statistically significant.
State a “real world” conclusion.

Based on our decision in step 4, we will write a sentence or two concerning our decision in relation to the original research question.

Example 8.9 (1 Sample Proportion Normal Approximation Method - Male Babies)

Video 8.4: 1 Sample Proportion Normal Approximation Method - Male Babies Example

Example 8.10 (Handedness) Research Question: Are more than 80% of American’s right-handed?

In a sample of 100 Americans, 87 were right-handed.

Check assumptions and write hypotheses.

$np_0 = 100(0.80)=80$

$n(1-p_0) = 100 (1-0.80) = 20$

Both $np_0$ and $n(1-p_0)$ are at least 10 so we can use the normal approximation method.

This is a right-tailed test because we want to know if the proportion is greater than 0.80.

$H_{0}\colon p=0.80$
$H_{a}\colon p>0.80$
Calculate the test statistic. (Equation 8.1)

\[\widehat{p}=\dfrac{87}{100}=0.87,\ p_{0}=0.80,\ n=100\]

\[ z= \dfrac{\widehat{p}- p_0 }{\sqrt{\dfrac{p_0 (1- p_0)}{n}}}= \dfrac{0.87-0.80}{\sqrt{\dfrac{0.80 (1-0.80)}{100}}}=1.75 \]

Our z-test statistic is 1.75.
Determine the p-value associated with the test statistic.

This is a right-tailed test so we need to find the area to the right of the test statistic, $z=1.75$, on the z-distribution.

Using Minitab, we find the probability $P(z\geq1.75)=0.0400592$ which may be rounded to $\text{p-value}=0.0401$.

Distribution plot of Density vs X - Normal, Mean=0, StDev=1 — Fig 8.3

Make a decision.

$p\leq .05$, therefore our decision is to reject the null hypothesis.
State a “real world” conclusion.

Yes, there is convincing evidence to state that more than 80% of all Americans are right-handed.

Example 8.11 (Ice Cream) Research Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%?

In a sample of 50 customers, 60% preferred chocolate over vanilla.

Check assumptions and write hypotheses.

$np_0 = 50(0.80) = 40$

$n(1-p_0)=50(1-0.80) = 10$

Both $np_0$ and $n(1-p_0)$ are at least 10. We can use the normal approximation method.

This is a left-tailed test because we want to know if the proportion is less than 0.80.

$H_{0}\colon p=0.80$
$H_{a}\colon p<0.80$
Calculate the test statistic. (Equation 8.1)

$\widehat{p}=0.60$, $p_{0}=0.80$, $n=50$

\[ z= \dfrac{\widehat{p}- p_0 }{\sqrt{\dfrac{p_0 (1- p_0)}{n}}}= \dfrac{0.60-0.80}{\sqrt{\dfrac{0.80 (1-0.80)}{50}}}=-3.536 \]

Our z-test statistic is -3.536.
Determine the p-value.

This is a left-tailed test so we need to find the area to the right of our test statistic, $z=-3.536$.

Fig 8.4

From the Minitab output above, the p-value is 0.0002031.
Make a decision.

$p \leq.05$, therefore our decision is to reject the null hypothesis.
State a “real world” conclusion.

Yes, there is convincing evidence that the percentage of all Creamery customers who prefer chocolate ice cream over vanilla is less than 80%.

Example 8.12 (Overweight Citizens) According to the Center for Disease Control (CDC), the percent of adults 20 years of age and over in the United States who are overweight is 69.0% (see NCHS FastStats: Obesity and Overweight). One city’s council wants to know if the proportion of overweight citizens in their city is different from this known national proportion. They take a random sample of 150 adults 20 years of age or older in their city and find that 98 are classified as overweight. Let’s use the five-step hypothesis testing procedure to determine if there is convincing evidence that the proportion in this city is different from the known national proportion.

Check assumptions and write hypotheses.

$np_0 =150 (0.690)=103.5$

$n (1-p_0) =150 (1-0.690)=46.5$

Both $n p_0$ and $n (1-p_0)$ are at least 10, this assumption has been met.

Research question: Is this city’s proportion of overweight individuals different from 0.690?

This is a non-directional test because our question states that we are looking for a difference as opposed to a specific direction. This will be a two-tailed test.

$H_{0}\colon p=0.690$
$H_{a}\colon p\neq 0.690$
Calculate the test statistic. (Equation 8.1)

\[ \widehat{p}=\dfrac{98}{150}=.653 \]

\[ z =\dfrac{0.653- 0.690 }{\sqrt{\dfrac{0.690 (1- 0.690)}{150}}} = -0.980 \]

Our test statistic is $z=-0.980$.
Determine the p-value.

This is a non-directional (i.e., two-tailed) test, so we need to find the area under the z-distribution that is more extreme than $z=-0.980$.

In Minitab, we find the proportion of a normal curve beyond $\pm0.980$:

Fig 8.5

\[ \text{p-value}=0.163543+0.163543=0.327086 \]
Make a decision.

$p>\alpha$, therefore we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not sufficient evidence to state that the proportion of citizens of this city who are overweight is different from the national proportion of 0.690.

Minitab: Hypothesis Tests for One Proportion

A hypothesis test for one proportion can be conducted in Minitab. This can be done using raw data or summarized data.

If you have a data file with every individual’s observation, then you have raw data.
If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data.

The next two examples will show you how to use Minitab to conduct this analysis using either raw data or summarized data.

Note that the default method for constructing the sampling distribution in Minitab is to use the exact method. If $np_0 \geq 10$ and $n(1-p_0) \geq 10$ then you will need to change this to the normal approximation method. This must be done manually. Minitab will use the method that you select, it will not check assumptions for you!

Minitab: 1 Proportion z-Test, Raw Data

If you have data in a Minitab worksheet, then you have what we call “raw data.” This is in contrast to “summarized data” which you’ll see in the next section.

In order to use the normal approximation method both $np_0 \geq 10$ and $n(1-p_0) \geq 10$. Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. This must be checked manually. Minitab will not check assumptions for you.

Conduct a one-proportion hypothesis test with raw data in Minitab:

Select Stat > Basic Statistics > 1 Proportion.
Select the variable containing the raw data.
Select Perform hypothesis test.
Enter the null-hypothesis value.
Select the Options tab.
Select the appropriate alternative hypothesis.
Under Method, select Normal approximation.
Select OK.

Example 8.13 (Minitab: Conducting a One Sample Proportion z-Test: Raw Data) The data for this example come from the class survey, which includes a sample of 226 students.

Research question: Is the proportion of students who are male different from 0.50?

Check assumptions and write hypotheses.

Here, $n=226$ and $p_0=0.50$. The expected numbers of successes and failures under the null hypothesis are

\[np_0=226(0.50)=113\]

and

\[n(1-p_0)=226(1-0.50)=113.\]

Both expected counts are at least 10. Therefore, the assumptions are satisfied, and we can use the normal approximation method.

$H_0\colon p = 0.50$

$H_a\colon p \ne 0.50$
Calculate the test statistic.

Use Minitab to conduct the one-sample proportion z-test with raw data:
1. Open the Minitab file: (class_survey.mpx).
2. Select Stat > Basic Statistics > 1 Proportion.
3. Select One or more samples, each in a column from the dropdown.
4. Double-click the variable Biological Sex to insert it into the box.
5. Check Perform hypothesis test and enter 0.50 in the Hypothesized proportion box.
6. Select Options.
7. Use the default alternative hypothesis, Proportion ≠ hypothesized proportion.
8. Use the default confidence level of 95.
9. Select Normal approximation method.
10. Select OK, then OK.
This should result in the following output:

Method

Event: Biological Sex = Male
p: proportion where Biological Sex = Male
Normal approximation is used for this analysis.

Descriptive Statistics

N Event Sample p 95% CI for p

226 99 0.438053 (0.373368, 0.502738)

Test

Null hypothesis H₀: p = 0.5
Alternative hypothesis H₁: p ≠ 0.5

Z-Value P-Value

-1.86 0.063

From the Minitab output, $z = -1.86$.
Determine the p-value.

From the Minitab output, $p = 0.063$.
Make a decision.

$p > \alpha$, fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence that the proportion of all students in the population who are male is different from 0.50.

N	Event	Sample p	95% CI for p
226	99	0.438053	(0.373368, 0.502738)

Z-Value	P-Value
-1.86	0.063

Minitab: 1 Proportion z-Test, Summarized Data

Use the summarized-data procedure when the individual observations are not available but the number of events and total number of trials are known. As with raw data, check the normal approximation assumptions before selecting the analysis method in Minitab.

Conduct a one-proportion hypothesis test with summarized data in Minitab:

Select Stat > Basic Statistics > 1 Proportion.
From the dropdown menu, select Summarized data.
Enter the summarized data.
Select Perform hypothesis test.
Enter the null-hypothesis value.
Select the Options tab.
Select the appropriate alternative hypothesis.
Under Method, select Normal approximation.
Select OK.

Example 8.14 (Minitab: Conducting a One Sample Proportion z-Test: Summarized Data) The following example uses a scenario in which we want to know if the proportion of college women who think they are overweight is less than 40%. We collect data from a random sample of 129 college women and 37 said that they think they are overweight.

Research question: Is the proportion of college women who think they are overweight less than 0.40?

Check assumptions and write hypotheses.

Here, $n=129$ and $p_0=0.40$. The expected numbers of successes and failures under the null hypothesis are

\[np_0=129(0.40)=51.6\]

and

\[n(1-p_0)=129(1-0.40)=77.4.\]

Both expected counts are at least 10. Therefore, the assumptions are satisfied, and we can use the normal approximation method.

$H_0\colon p = 0.40$

$H_a\colon p < 0.40$
Calculate the test statistic.

Use Minitab to conduct the one-sample proportion z-test with summarized data:
1. Select Stat > Basic Statistics > 1 Proportion.
2. Select Summarized data from the dropdown.
3. Enter 37 for the number of events and 129 for the number of trials.
4. Check Perform hypothesis test and enter 0.40 in the Hypothesized proportion box.
5. Select Options.
6. Set the alternative hypothesis to Proportion < hypothesized proportion.
7. Use the default confidence level of 95.
8. Select Normal approximation method.
9. Select OK, then OK.
This should result in the following output:

Method

Event: Event proportion
Normal approximation is used for this analysis.

Descriptive Statistics

N Event Sample p 95% Upper Bound for p

129 37 0.286822 0.352321

Test

Null hypothesis H₀: p = 0.4
Alternative hypothesis H₁: p < 0.4

Z-Value P-Value

-2.62 0.004

From the Minitab output, $z=-2.62$.
Determine the p-value.

From the Minitab output, $p=0.004$.
Make a decision.

$p \leq \alpha$, reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence that the proportion of women in the population who think they are overweight is less than 40%.

N	Event	Sample p	95% Upper Bound for p
129	37	0.286822	0.352321

Z-Value	P-Value
-2.62	0.004

Example 8.15 (Minitab: Normal Approx. Method (Gym membership)) Research question: Are less than 50% of all individuals with a membership at one gym female?

A simple random sample of 60 individuals with a membership at one gym was collected. Each individual’s biological sex was recorded. There were 24 females.

Check assumptions and write hypotheses.

$np_0=60(0.50)=30$ and $n(1-p_0)=60(1-0.50)=30$. Both values are at least 10, so we can use the normal approximation method.

$H_0\colon p = 0.50$
$H_a\colon p < 0.50$
Calculate the test statistic.

Use Minitab to conduct a one-sample proportion z-test with summarized data:
1. Select Stat > Basic Statistics > 1 Proportion.
2. Select Summarized data from the dropdown.
3. Enter 24 for the number of events and 60 for the number of trials.
4. Select Perform hypothesis test and enter 0.50 for the hypothesized proportion.
5. Select Options.
6. For the alternative hypothesis, select Proportion < hypothesized proportion.
7. Use the default confidence level of 95.
8. Select Normal approximation method.
9. Select OK, then OK.
This should result in the following output:

Method

Event: Event proportion
Normal approximation is used for this analysis.

Descriptive Statistics

N Event Sample p 95% Upper Bound for p

60 24 0.400000 0.504030

Test

Null hypothesis $H_0: p=0.50$

Alternative hypothesis $H_1: p<0.50$

Z-Value P-Value

-1.55 0.061

From the Minitab output, the test statistic is $z=-1.55$.
Determine the p-value.

From the output, $p=0.061$.
Make a decision.

Because $p \geq \alpha$, we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not enough evidence to conclude that less than 50% of all members at this gym are female.

N	Event	Sample p	95% Upper Bound for p
60	24	0.400000	0.504030

Null hypothesis	\(H_0: p=0.50\)
Alternative hypothesis	\(H_1: p<0.50\)

Z-Value	P-Value
-1.55	0.061

8.2 One Sample Mean

One sample mean tests are covered in Section 6.2 of the Lock⁵ textbook.

Concerning one sample mean, the Central Limit Theorem states that if the sample size is large, then the distribution of sample means will be approximately normally distributed with a standard deviation (i.e., standard error) equal to $\frac{\sigma}{\sqrt n}$. In this course, a “large” sample size will be defined as one where $n \ge 30$.

When constructing confidence interval and conducting hypothesis tests we often do not know the value of $\sigma$. In those cases, $\sigma$ may be estimated using the sample standard deviation ($s$). When we are using $s$ to estimate $\sigma$ our sampling distribution will not follow a z-distribution exactly. Instead, we use what is known as the t-distribution. Like the z-distribution, the t-distribution is symmetrical. The difference is that its height varies depending on the sample size. By doing so, the distribution becomes more conservative for smaller sample sizes to account for some error that may occur from estimating $\sigma$ with $s$ from a small sample. As $n$ approaches infinity ($\infty$) the t-distribution approaches the standard normal distribution. The next section compares the z- and t-distributions.

When constructing confidence intervals and conducting hypothesis tests we will usually be using the t-distribution when working with one mean. The only exception would be in cases where $\sigma$ is known. This scenario is most common in the fields of education and psychology where some tests are normed to have a certain $\mu$ and $\sigma$. In those cases, the z-distribution can be used.

In terms of language, all of these tests could be called “single sample mean tests” or “one sample mean tests.” We could also specify the sampling distribution by using the term “single sample mean t-test” or “single sample mean z-test.”

The flow chart below may help you in determining which method should be used when constructing a sampling distribution for one sample mean.

Identify when z- and t-distributions should be used.

Is the population known to be normally distributed?
Is the population standard deviation known?
Is the sample size at least 30?

Text alternative of Figure 8.6

Is the population known to be normally distributed?
- Yes:
  - Is the population standard deviation known?
    - Yes: z-distribution.
    - No: t-distribution.
- No:
  - Is the sample size at least 30?
    - Yes: t-distribution.
    - No: Bootstrap/Randomization.

8.2.1 t-Distribution

The height of the t-distribution is determined by the number of degrees of freedom (df). For a one sample mean test, $df=n-1$.

The first plot below compares the standard normal distribution (i.e., z-distribution) to a t-distribution. The solid blue line is the standard normal distribution and the dashed red line is a t-distribution with 2 degrees of freedom. Here, the tails of the t-distribution are higher than the tails of the normal distribution.

A plot showing the z-distribution compared to a t distribution with df=2. — Fig 8.7

If you think about the area under the curve, the higher tails mean that more area will fall in the tails. For example, as seen in the following two plots, $P(z>2.00)=0.0227501$ while $P(t_{df=2}>2.00)=0.0917517$.

Standard normal (i.e., z) distribution showing the area above z=2 — Fig 8.8

Probability distribution plot showing the area greater than t=2 on a distribution with 2 degrees of freedom — Fig 8.9

The next plot compares the standard normal distribution to a t distribution with 10 degrees of freedom. Notice that the two distributions are becoming more similar as the sample size increases.

Plot comparing the z-distribution to a t distribution with 10 degrees of freedom — Fig 8.10

The next plot compares the standard normal distribution to a t distribution with 30 degrees of freedom.

Plot comparing the standard normal distribution to a t distribution with 30 degrees of freedom — Fig 8.11

In the final graph, the standard normal distribution is compared to a t distribution with 500 degrees of freedom. Here, the two distributions are nearly identical. As the degrees of freedom approach infinity, the t distribution approaches (i.e., becomes more similar to) the standard normal distribution.

Plot comparing the standard normal distribution to a t distribution with 500 degrees of freedom — Fig 8.12

Minitab: Using t-Distribution Probability Plots

The procedures for constructing t distributions in Minitab are similar to those for constructing z-distributions. We can construct a probability distribution plot to find the $t^{*}$ multiplier when constructing a confidence interval. And, we can construct a plot to find the p-value when conducting a hypothesis test.

Find a $t^*$ multiplier in Minitab:

Select Graph > Probability Distribution Plot.
Select One Curve, then select View Probability.
Select OK.
Change the Distribution to t.
Enter the Degrees of freedom.
Select Options.
Select A specified probability.
Select Equal tails.
Under Probability, enter the combined proportion in the two tails. For example, enter 0.10 for a 90% confidence interval.
Select OK, then select OK again.

Find a p-value from a $t$-test statistic in Minitab:

Select Graph > Probability Distribution Plot.
Select One Curve, then select View Probability.
Select OK.
Change the Distribution to t.
Enter the Degrees of freedom.
Select Options.
Select A specified x value.
Select the region that matches the alternative hypothesis:
- Left tail for a left-tailed test.
- Right tail for a right-tailed test.
- Equal tails for a two-tailed test.
Under X value, enter the $t$-test statistic.
Select OK, then select OK again.

8.2.2 Confidence Intervals

Confidence intervals are used to estimate unknown population parameters. Because the population standard deviation ($\sigma$) will almost always be unknown in situations in which we are constructing confidence intervals for means, the t-distribution is used to estimate the sampling distribution. The following section will show you how to construct a confidence interval for a population mean using formulas and using Minitab. Similar to how we computed necessary minimum sample sizes for confidence intervals for proportions, we will also compute the necessary minimum sample size for constructing a confidence interval for a mean.

Formulas

Earlier in this lesson we considered confidence intervals for proportions and the multiplier in our intervals was a value from the standard normal (i.e., $z$) distribution. But, what if our variable of interest is a quantitative variable, and we want to estimate a population mean?

We apply similar techniques when constructing a confidence interval for a mean, but now we are interested in estimating the population mean ($\mu$) by using the sample statistic ($\bar{x}$) and the multiplier is a $t$ value. Similar to the $z$ values that you used as the multiplier for constructing confidence intervals for population proportions, here you will use $t$ values as the multipliers. Because $t$ values vary depending on the number of degrees of freedom (df), you will need to use statistical software to look up the appropriate $t$ value for each confidence interval that you construct. The degrees of freedom will be based on the sample size. Since we are working with one sample here, $df=n-1$.

Example 8.16 (Minitab: Finding $t^*$ Multipliers) To find the $t^{*}$ multiplier for a 98% confidence interval with 15 degrees of freedom:

In Minitab, select Graph > Probability Distribution Plot > View Probability.
Change the Distribution to t.
Enter 15 for the Degrees of freedom.
Select Options.
Choose A specified probability.
Select Equal tails.
For Probability, enter 0.02 (if there is 0.98 in the middle, then 0.02 is split equally between the left and right tails).

This should result in an output similar to the output below. Note that your results may be slightly different due to random sampling variation.

T Distribution, 98% CI, DF=15 — Fig 8.13

Let’s review some of symbols and equations that we learned in previous lessons:

Sample size: $n$
Population mean: $\mu = \dfrac{\sum X}{N}$
Sample mean: $\bar{x} = \dfrac{\sum x}{n}$
Standard error of the mean: $\operatorname{SE} = \dfrac{s}{\sqrt{n}}$
Multiplier: $t^*$
Degrees of freedom (one group): $df = n - 1$

Recall the general form for a confidence interval:

General Form of Confidence Interval

\[ \text{sample statistic}\pm\underbrace{(\text{multiplier})\ (\text{standard error})}_{\textbf{margin of error}} \]

When constructing a confidence interval for a population mean the point estimate is the sample mean, $\bar{x}$. The multiplier is taken from a t-distribution. And, the standard error is equal to $\frac{s}{\sqrt{n}}$.

Confidence Interval for a Population Mean

\[ \underbrace{\bar{x}}_{\text{sample statistic}} \pm \overbrace{t^{*}}^{\text{multiplier}} \underbrace{ \dfrac{s}{\sqrt{n}}}_{\text{standard error}} \]

The following examples walk through constructing confidence intervals for population means by hand. Then, you will learn how to compute confidence intervals using Minitab.

Example 8.17 (MLB Age) In a sample of 30 current MLB pitchers, the mean age was 28 years with a standard deviation of 4.4 years. Construct a 95% confidence interval to estimate the mean age of all current MLB pitchers.

This is what we know: $n=30$, $\bar{x}=28$, and $s=4.4$.

In order to compute the confidence interval for $\mu$ we will need the t multiplier and the standard error ($\frac{s}{\sqrt{n}}$).

\[ df=n-1=30-1=29 \]

t Distribution showing the multipliers for a 95% confidence interval given 21 degrees of freedom — Fig 8.14

For a 95% confidence interval with 29 degrees of freedom, $t^{*}=2.045$

\[ \text{SE}=\dfrac{s}{\sqrt{n}}=\dfrac{4.4}{\sqrt{30}}=0.803 \]

Thus, our confidence interval for $\mu$ is:

\[ 28\pm 2.045(0.803)=28\pm1.643=[26.357,29.643] \]

We are 95% confident that the population mean age is between 26.357 and 29.643.

Example 8.18 (Sleep Deprivation) In a class survey, students were asked how many hours they sleep per night. In the sample of 22 students, the mean was 5.77 hours with a standard deviation of 1.572 hours. That distribution was approximately normal. Let’s construct a 95% confidence interval for the mean number of hours slept per night in the population from which this sample was drawn.

This is what we know: $n=22$, $\bar{x}=5.77$, and $s=1.572$.

In order to compute the confidence interval for $\mu$ we will need the t multiplier and the standard error ($\frac{s}{\sqrt{n}}$).

\[ df=n-1=22-1=21 \]

For a 95% confidence interval with 21 degrees of freedom, $t^{*}=2.080$

\[ \text{SE}=\frac{s}{\sqrt{n}}=\frac{1.572}{\sqrt{22}}=0.335 \]

Thus, our confidence interval for $\mu$ is:

\[ 5.77\pm 2.080(0.335)=5.77\pm0.697=[5.073, 6.467] \]

We are 95% confident that the population mean is between 5.073 and 6.467 hours.

What if we wanted to be more conservative and construct a 99% confidence interval?

The only thing that would change is our multiplier. Now, $t^{*}=2.831$.

t Distribution showing the multipliers for a 99% confidence interval given 21 degrees of freedom — Fig 8.16

\[ 5.77\pm 2.831(0.335)=5.77\pm0.948=[4.822, 6.718] \]

We are 99% confident that the population mean is between 4.822 and 6.718 hours.

Example 8.19 (Milk) A study of 66,831 dairy cows found that the mean milk yield was 12.5 kg per milking with a standard deviation of 4.3 kg per milking (data from Berry, et al., 2013). Construct a 95% confidence interval for the average milk yield in the population.

First, let’s compute the standard error:

\[ \text{SE}=\dfrac{s}{\sqrt{n}}=\dfrac{4.3}{\sqrt{66831}}=0.0166 \]

The standard error is small because the sample size is very large.

Next, let’s find the $t^*$ multiplier:

\[ df=66831-1=66830 \]

Distribution Plot - T, DF=66830 — Fig 8.17

\[ t^{*}=1.960 \]

Now, we can construct our 95% confidence interval:

95% CI: $12.5\pm1.960(0.017)=12.5\pm0.033=[12.467,\;12.533]$

We are 95% confident that the mean milk yield in the population is between 12.467 and 12.533 kg per milking.

Example 8.20 (Minitab: Confidence Interval of a Mean) Here you will learn how to use Minitab to construct a confidence interval for a mean. The procedure is similar to the one that you learned earlier in this lesson for constructing a confidence interval for a proportion. The following example walks through this procedure when data are in a Minitab work.

To create a 95% confidence interval of mean height in Minitab:

Open the data set: (fall2016stdata.csv).
In Minitab, select Stat > Basic Statistics > 1-sample t.
In this case, we have our data in the Minitab worksheet, so we will use the default One or more samples, each in a column.
Insert the variable Height in the box on the left into the box on the right.
Select Options.
The default Confidence level is 95.
Select OK, then OK.

This should result in the following output:

Descriptive Statistics

N	Mean	StDev	SE Mean	95% CI for $\mu$
525	67.009	4.462	0.195	(66.627, 67.392)

$\mu$: mean of Height

What if we have summarized data and not data in a Minitab worksheet?

If you do not have a Minitab worksheet filled with data concerning individuals, but instead have summarized data (e.g., the values of $s$, $\bar{x}$, and $n$), you would skip step 1 above and in step 3 you would select Summarized data.

Try It!

Estimating the average MLB Pitcher’s age In a sample of 30 current MLB pitchers, the mean age was 28 years with a standard deviation of 4.4 years. Construct a 95% confidence interval to estimate the mean age of all current MLB pitchers.
We know that n = 30, $\bar{x}=28$, and s = 4.4.

To create a 95% confidence interval of mean age in Minitab:
1. In Minitab, select Stat > Basic Statistics > 1-sample t.
2. In this case we have summarized data so select Summarized Data from the dropdown
3. Enter 30 for the sample size, 28 for the sample mean, and 4.4 for the standard deviation.
4. Select Options.
5. The default Confidence level is 95.
6. Select OK, then OK.
This should result in the following output:

Descriptive Statistics

N Mean StDev SE Mean 95% CI for $\mu$

30 28.000 4.400 0.803 (26.357, 29.643)

$\mu$: population mean of sample

We are 95% confident that the population mean age is between 26.357 and 29.643 years.
Coffee Sales (Data in Column) For 48 days data concerning sales were collected from one student-run cafe. Construct a 95% confidence interval for the mean number of coffees sold per day.
To create a 95% confidence interval of mean number of coffees sold per day in Minitab:
1. Open the file: (cafedata.mpx).
2. In Minitab, select Stat > Basic Statistics > 1-sample t.
3. In this case the data is in a worksheet, so use One or more samples, each in a column.
4. Select the variable Coffees.
5. Select Options.
6. The default Confidence level is 95.
7. Select OK, then OK.
This should result in the following output:

Descriptive Statistics

N Mean StDev SE Mean 95% CI for $\mu$

47 21.51 11.08 1.62 (18.26, 24.76)

$\mu$: population mean of Coffees

We are 95% confident that the population mean number of coffees sold per day is between 18.26 and 24.76.

N	Mean	StDev	SE Mean	95% CI for \(\mu\)
30	28.000	4.400	0.803	(26.357, 29.643)

N	Mean	StDev	SE Mean	95% CI for \(\mu\)
47	21.51	11.08	1.62	(18.26, 24.76)

8.2.3 Computing Necessary Sample Size

Calculating the sample size necessary for estimating a population mean with a given margin of error and level of confidence is similar to that for estimating a population proportion. However, since the t-distribution is not as “neat” as the standard normal distribution, the process can be iterative. (Recall, the shape of the t-distribution is different for each degree of freedom). This means that we would solve, reset, solve, reset, etc. until we reached a conclusion. Yet, we can avoid this iterative process if we employ an approximate method based on t-distribution approaching the standard normal distribution as the sample size increases. This approximate method invokes the following formula:

Finding the Sample Size for Estimating a Population Mean

\[ n=\dfrac{z^{2}\widetilde{\sigma}^{2}}{M^{2}}=\left ( \dfrac{z\widetilde{\sigma}}{M} \right )^2 \]

$z$ = z multiplier for given confidence level
$\widetilde{\sigma}$ = estimated population standard deviation
$M$ = margin of error

The sample standard deviation may be estimated on the basis of prior research studies.

Example 8.21 (Estimating IQ) A team of researchers wants to estimate the mean IQ of students enrolled at one prestigious university. Previous research studies have examined samples of students from other similar universities and usually find results around $\bar{x}=120$ and $s=10$. In order to construct a 90% confidence interval with a margin of error of $\pm2$ IQ points, what sample size should be obtained?

As shown in the probability distribution plot below, the z value associated with a 90% confidence interval is 1.645.

Standard normal distribution showing the z multipliers for a 90% confidence interval — Fig 8.18

The estimated standard deviation is given to be 10 and the desired margin of error is given to be 2.

\[ n=\dfrac{z^{2}\widetilde{\sigma}^{2}}{M^{2}}=\dfrac{1.645^{2}(10^{2})}{2^{2}}=67.615 \]

We round up to 68. The research team should attempt to obtain a sample of at least 68 individuals.

Example 8.22 (Video: Age)

Video 8.5: STAT 200 Estimating Necessary Sample Size for CI of Mu (Age Example)

8.2.4 Hypothesis Testing

In this section, we will be comparing one sample mean to one known or hypothesized population value. In Lesson 5 you learned how to conduct randomization tests. Here, you will learn how to conduct a one sample mean t-test and a one sample mean z-test. The t-distribution is used to estimate the sampling distribution when the sample size is large (at least 30) or when the population is known to be normally distributed (but $\sigma$ is unknown). The z-distribution is used on rare occasions when the population is normal and the population standard deviation is known. Note that for this course the one sample mean z-test is optional; it used only in specific cases where the population is known to be normally distributed and when the population standard deviation ($\sigma$) is known. The most commonly used one sample mean test is the “one sample mean t-test” which is also known as a “single sample mean t-test.”

Text alternative of Figure 8.19

Is the population known to be normally distributed?
- Yes:
  - Is the population standard deviation known?
    - Yes: z-distribution.
    - No: t-distribution.
- No:
  - Is the sample size at least 30?
    - Yes: t-distribution.
    - No: Bootstrap/Randomization.

One Sample Mean t-Test, Formulas

Five-Step Hypothesis Testing Procedure

Check assumptions and write hypotheses.

Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large ($\ge\ 30$) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are:

Research Question	Is the mean different from $\mu_{0}$?	Is the mean greater than $\mu_{0}$?	Is the mean less than $\mu_{0}$?
Null Hypothesis, $H_{0}$	$\mu=\mu_{0}$	$\mu=\mu_{0}$	$\mu=\mu_{0}$
Alternative Hypothesis, $H_{a}$	$\mu\neq \mu_{0}$	$\mu> \mu_{0}$	$\mu<\mu_{0}$
Type of Hypothesis Test	Two-tailed, non-directional	Right-tailed, directional	Left-tailed, directional

where $\mu_{0}$ is the hypothesized population mean.

Calculate the test statistic.

For the test of one group mean we will be using a t-test statistic:

Test Statistic: One Group Mean

\[ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}} \tag{8.2}\]

$\bar{x}$ = sample mean
$\mu_{0}$ = hypothesized population mean
$s$ = sample standard deviation
$n$ = sample size

Note that structure of this formula is similar to the general formula for a test statistic:

\[ \dfrac{\text{sample statistic}-\text{null value}}{\text{standard error}} \]
Determine the p-value.

When testing hypotheses about a mean or mean difference, a t-distribution is used to find the $p$-value. These t-distributions are indexed by a quantity called degrees of freedom, calculated as $df = n – 1$ for the situation involving a test of one mean or test of mean difference. The $p$-value can be found using Minitab.
Make a decision.

If $p \leq \alpha$ reject the null hypothesis.

If $p>\alpha$ fail to reject the null hypothesis.
State a “real world” conclusion.

Based on your decision in Step 4, write a conclusion in terms of the original research question.

Next, you will go through a few examples, and then you will have the opportunity to do two on your own.

Example 8.24 (Book Costs) Research question: Does the average Penn State student spend more than $300 each semester on textbooks?

In a sample of 226 Penn State students, the mean cost of a student’s textbooks was $344 with a standard deviation of $106.

Video 8.7: STAT 200 Single Sample Mean t Hypothesis Test Example: Textbook Spending

Example 8.25 (Pulse Rate) A research study measured the pulse rates of 57 college men and found a mean pulse rate of 70.4211 beats per minute with a standard deviation of 9.9480 beats per minute. Researchers want to know if the mean pulse rate for all college men is different from the current standard of 72 beats per minute.

Check assumptions and write hypotheses.

Pulse rates are quantitative. The sampling distribution will be approximately normally distributed because $n \ge 30$.

This is a two-tailed test because we want to know if the mean pulse rate is different from 72.

$H_{0}:\mu=72$
$H_{a}: \mu\neq 72$
Calculate the test statistic.(Equation 8.2)

\[ \bar{x}=70.4211, \mu_0=72, s=9.9480, n=57 \] \[ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{70.4211-72}{\dfrac{9.9480}{\sqrt{57}}}=-1.198 \]

Our t-test statistic is -1.198.
Determine the p-value.

\[ df=n-1=57-1=56 \]

Fig 8.20

\[ p=0.117981+0.117981=0.235962 \]

Given that the null hypothesis is true and $\mu=72$, the probability of taking a random sample of $n=57$ and finding a sample mean this or more extremely different is 0.235962. This is our p-value.
Make a decision.

$p>.05$, therefore we fail to reject the null hypothesis.
State a “real world” conclusion.

There is not sufficient evidence to state that the mean pulse of college men is different from 72.

Example 8.26 (Coffee) In the population of Americans who drink coffee, the average daily consumption is 3 cups per day. A university wants to know if their students tend to drink more coffee than the national average. They ask a random sample of 50 students how many cups of coffee they drink each day and found $\bar{x}=3.8$ and $s=1.5$. Do they have convincing evidence that their students drink more than the national average?

Check assumption and write hypotheses.

Amount of coffee consumed is a quantitative variable. We are given that random sampling methods were employed. Because $n \ge 30$, we can approximate the sampling distribution using a t distribution.

This is a right-tailed test because we want to know if the mean in the sample is greater than the national average.

$H_{0}:\mu= 3$
$H_{a}:\mu>3$
Calculate the test statistic. (Equation 8.2)

\[ \bar{x}=3.8, \mu_0=3, s=1.5, n=50 \]

\[ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{3.8-3}{\dfrac{1.5}{\sqrt{50}}}=3.771 \]

Our t-test statistic is 3.771
Determine the p-value.

$df=n-1=50-1=49$

Fig 8.21

Using Minitab, we can find that $P(t > 3.771) =0.0002191$.

p-value = 0.0002191

If $\mu=3$, then the probability of taking a random sample of $n=50$ and finding $\bar{x} \geq 3$ is 0.0002191
Make a decision.

$p\leq.05$, therefore we reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence to state the mean number of coffees consumed in the population of all students at this university is greater than 3.

Example 8.27 (Transportation Costs) According to CNN, in 2011, the average American spent $16,803 on housing. A suburban community wants to know if their residents spent less than this national average. In a survey of 30 randomly selected residents, they found that they spent an annual average of $15,800 with a standard deviation of $2,600.

Check assumptions and write hypotheses.

Housing costs are quantitative. Because $n \ge 30$, the sampling distribution can be approximated using the t-distribution.

This is a left-tailed test because we want to know if residents of this community spent less than the national average.

$H_{0}:\mu=16803$
$H_{a}:\mu<16803$
Calculate the test statistic. (Equation 8.2)

\[ \bar{x}=15800, \mu_0=16803, s=2600, n=30 \]

\[ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{15800-16803}{\dfrac{2600}{\sqrt{30}}}=-2.113 \]

Our t-test statistics is -2.113.
Determine the p-value.

\[ df=n-1=30-1=29 \]

This is a left-tailed test so we want to know the probability of $t < -2.113$.

Fig 8.22

Using Minitab we can find that $p=0.0216634$.
Make a decision.

$p\leq .05$, therefore we reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence to state that on average residents of this community spent less than the national average on housing in 2011.

Minitab: One Sample Mean t-Tests

A hypothesis test for one group mean can be conducted in Minitab using raw data or summarized data.

If you have a data file with every individual’s observation, then you have raw data.
If you do not have each individual’s observation, but rather have the sample mean, sample standard deviation, and sample size, then you have summarized data.

Conduct a one-sample t hypothesis test with raw data in Minitab:

Select Stat > Basic Statistics > 1-Sample t.
Select the variable containing the raw data.
Select Perform hypothesis test.
Enter the null-hypothesis value.
Select the Options tab.
Select the appropriate alternative hypothesis.
Select OK.

Conduct a one-sample t hypothesis test with summarized data in Minitab:

Select Stat > Basic Statistics > 1-Sample t.
From the dropdown menu, select Summarized data.
Enter the sample size, sample mean, and sample standard deviation.
Select Perform hypothesis test.
Enter the null-hypothesis value.
Select the Options tab.
Select the appropriate alternative hypothesis.
Select OK.

The next two examples will show you how to use Minitab to conduct a one-sample mean t-test using either raw data or summarized data. There is also one example of using Minitab to conduct a one-sample mean z-test which is only performed if the population is known to be normally distributed and the population standard deviation ($\sigma$) is available.

Example 8.28 (Minitab: One Sample Mean t-Test Using Raw Data) Research question: Is the mean GPA in the population different from 3.0?

The GPAs of 226 students are available.

Check assumptions and write hypotheses.

We do not know whether the population is normally distributed. However, the sample size is large ($n=226 \ge 30$), so the sampling distribution of the sample mean is approximately normal and we can perform a one-sample mean t-test.

$H_0\colon \mu = 3.0$
$H_a\colon \mu \ne 3.0$
Calculate the test statistic.

Use Minitab to conduct the one-sample mean t-test with raw data:
1. Open the Minitab file: (class_survey.mpx).
2. Select Stat > Basic Statistics > 1-Sample t.
3. Select One or more samples, each in a column from the dropdown.
4. Insert the variable GPA into the Sample box.
5. Check Perform a hypothesis test.
6. Enter 3 for the Hypothesized mean.
7. Select Options.
8. Use the default alternative hypothesis, Mean ≠ hypothesized mean.
9. Use the default confidence level of 95.
10. Select OK, then OK.
This should result in the following output:

N Mean StDev SE Mean 95% CI for $\mu$

226 3.2311 0.5104 0.0340 (3.1642, 3.2980)

$\mu$: population mean of GPA

Test

Null hypothesis: $H_0\colon \mu = 3$
Alternative hypothesis: $H_1\colon \mu \ne 3$

T-Value P-Value

6.81 0.000

From the Minitab output, $t(225)=6.81$.
Determine the p-value.

Minitab reports the p-value as 0.000. We write this as $p<0.001$.
Make a decision.

$p \le \alpha$, so we reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence that the mean GPA in the population is different from 3.0.

N	Mean	StDev	SE Mean	95% CI for \(\mu\)
226	3.2311	0.5104	0.0340	(3.1642, 3.2980)

T-Value	P-Value
6.81	0.000

Example 8.29 (Minitab: One Sample Mean t-Test Using Summarized Data) Here we are testing $H_{a}\colon\mu\neq72$ and are given $n=35$, $\bar{x}=76.8$, and $s=11.62$.

Check assumptions and write hypotheses.

The shape of the population distribution is unknown. However, because $n \ge 30$, we can perform a one-sample mean t-test.

$H_0\colon \mu = 72$
$H_a\colon \mu \ne 72$
Calculate the test statistic.

Use Minitab to conduct the one-sample mean t-test with summarized data:
1. Select Stat > Basic Statistics > 1-Sample t.
2. Select Summarized data from the dropdown.
3. Enter 35 for the sample size, 76.8 for the sample mean, and 11.62 for the standard deviation.
4. Select Perform a hypothesis test.
5. Enter 72 for the hypothesized mean.
6. Select Options.
7. Use the default alternative hypothesis of Mean ≠ hypothesized mean.
8. Use the default confidence level of 95.
9. Select OK, then OK.
This should result in the following output:

Descriptive Statistics

N Mean StDev SE Mean 95% CI for $\mu$

35 76.80 11.62 1.96 (72.81, 80.79)

$\mu$: population mean of Sample

Test

Null hypothesis: $H_0\colon \mu = 72$
Alternative hypothesis: $H_1\colon \mu \ne 72$

T-Value P-Value

2.44 0.0199

From the Minitab output, the test statistic is $t(34)=2.44$.
Determine the p-value.

From the output, $p=0.0199$.
Make a decision.

Because $p \le \alpha$, we reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence that the population mean is different from 72.

N	Mean	StDev	SE Mean	95% CI for \(\mu\)
35	76.80	11.62	1.96	(72.81, 80.79)

T-Value	P-Value
2.44	0.0199

One Sample Mean z-Test (Optional)

A one sample mean z-test is used when the population is known to be normally distributed and when the population standard deviation ($\sigma$) is known. This most frequently occurs in the social sciences when standardized measures are used such as IQ, SAT, ACT, or GRE scores, for which the population parameters are known.

The formula for computing a z-test statistic for one sample mean is identical to that of computing a t-test statistic for one sample mean, except now the population standard deviation is known and can be used in computing the standard error.

z-Test Statistic: One Group Mean

\[ z=\dfrac{\bar{x}-\mu_0}{\dfrac{\sigma}{\sqrt{n}}} \]

$\bar{x}$ = sample mean
$\mu_{0}$ = hypothesized population mean
$s$ = sample standard deviation
$n$ = sample size

The other primary difference between the one sample mean t-test and the one sample mean z-test is the latter uses the standard normal distribution (i.e., z-distribution) in determining the p-value. Below are the directions for conducting a one sample mean z-test in Minitab.

Example 8.30 (Minitab: Performing a One Sample Mean z-Test) Research question: Are the IQ scores of students at one college-prep school above the national average?

Scores on one American IQ test are normed to have a mean of 100 and standard deviation of 15. In a simple random sample of 25 students at this school the mean was 110.

Check assumptions and write hypotheses.

The population is known to be normally distributed, and the population standard deviation is known to be 15. With these two conditions met, we can conduct a one-sample mean z-test.

$H_0\colon \mu = 100$
$H_a\colon \mu > 100$
Calculate the test statistic.

Use Minitab to conduct the one-sample mean z-test with summarized data:
1. Select Stat > Basic Statistics > 1-Sample Z.
2. Select Summarized data from the dropdown.
3. Enter 25 for the sample size, 110 for the sample mean, and 15 for the known standard deviation.
4. Check Perform a hypothesis test.
5. Enter 100 for the Hypothesized mean.
6. Select Options.
7. Set the alternative hypothesis to Mean > hypothesized mean.
8. Use the default confidence level of 95.
9. Select OK, then OK.
This should result in the following output:

Descriptive Statistics

N Mean SE Mean 95% Lower Bound for $\mu$

25 110.00 3.00 105.07

$\mu$: population mean of Sample
Known standard deviation = 15

Test

Null hypothesis: $H_0\colon \mu = 100$
Alternative hypothesis: $H_1\colon \mu > 100$

Z-Value P-Value

3.33 0.000

From the Minitab output, $z=3.33$.
Determine the p-value.

Minitab reports the p-value as 0.000. We write this as $p<0.001$.
Make a decision.

$p \le \alpha$, so we reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence that the mean IQ score of all students at this school is greater than 100.

N	Mean	SE Mean	95% Lower Bound for \(\mu\)
25	110.00	3.00	105.07

Z-Value	P-Value
3.33	0.000

8.3 Paired Means

In Lesson 1 we learned about independent samples and paired samples. When we have two independent samples, the observations in the two groups are unrelated to one another and are not matched in any meaningful way. We’ll learn how to compare the means of two independent groups in Lesson 9.

With paired samples, the observations in the two groups are matched in a meaningful way. These are also known as dependent samples. Most often this occurs when data are collected twice from the same participants, called repeated measures. For example, think of studying the effectiveness of a diet plan. You would weigh each participant prior to starting the diet and again following some time on the diet. Depending on how much weight they lost you would determine if the diet was effective. Paired data does not always need to involve two measurements on the same subject; it can also involve taking one measurement on each of two related subjects. For example, we may study husband-wife pairs, mother-son pairs, or pairs of twins.

In constructing a dependent samples confidence interval or conducting a dependent samples hypothesis test, the difference score is computed for each individual or pair. From there, the procedures are the same that you used for constructing confidence intervals and hypothesis tests for single sample means. As with one sample mean, if the sample size is at least 30, the sampling distribution for the difference in paired means can be approximated using a t-distribution.

In terms of symbols, the population parameter of interest is the mean difference in the population “$\mu_d$.” This is estimated using the mean difference in the sample “$\bar x_d.$”

8.3.1 Confidence Intervals

Recall the general form of a confidence interval…

\[ \text{sample statistic}\pm\underbrace{(\text{multiplier})\ (\text{standard error})}_{\textbf{margin of error}}. \]

The formula for constructing a confidence interval for the difference in paired means is almost identical to the formula for constructing a confidence interval for one mean. Note that the only change is the subscript d which stands for difference.

Confidence Interval for the Difference Between Two Paired Means

\[ \underbrace{\bar{x}_d}_{\text{sample statistic}} \pm \overbrace{t^*}^{\text{multiplier}} \underbrace{\left(\dfrac{s_d}{\sqrt{n}}\right)}_{\text{standard error}} \]

$t^*$ is the multiplier with $df = n-1$.

Example 8.31 (Change in Knowledge) An educational research study is designed so that participants complete a measure of demonstrated knowledge twice. The researcher wants to estimate the change in scores from the first to second administrations (i.e., pre- and post-test). Data are paired by participant. The researcher subtracted pre-test scores from the post test scores and found a mean increase of 6.560 with a standard deviation of 3.867 for $n=100$. She wants to construct a 95% confidence interval for the mean difference.

First, we’ll find the appropriate multiplier.

\[ df=n-1=100-1=99 \]

t distribution showing the multipliers for a 95% confidence interval given 99 degrees of freedom. — Fig 8.23

For a 95% confidence interval: $t_{df=99}=1.984$

\[ 6.560 \pm 1.984 \left(\frac{3.867}{\sqrt{100}}\right)=6.560 \pm 0.767=[5.793, 7.327] \]

We are 95% confident that the mean difference between post- and pre-test scores in the population is between 5.793 and 7.327.

Data from Zimmerman, W. A. (2015). Impact of Instructional Materials Eliciting Low and High Cognitive Load on Self-Efficacy and Demonstrated Knowledge (Unpublished doctoral dissertation). The Pennsylvania State University, University Park, PA.

Example 8.32 (Video: Difference in Exam Scores)

Video 8.8: STAT 200 95% CI for Mean Difference in Exam Scores

Minitab: Paired Means Confidence Interval

Construct a paired t confidence interval in Minitab:

Select Stat > Basic Statistics > Paired t.
Enter the appropriate columns under Sample 1 and Sample 2.
Select the Options tab.
Enter the appropriate confidence level.
Set the alternative hypothesis to not equal.
Select OK.

8.3.2 Hypothesis Testing

Below are the procedures for conducting a hypothesis test for two paired means. This is often referred to as a “paired means t-test,” “dependent means t-test,” or “matched pairs t-test.”

Check any necessary assumptions and write null and alternative hypotheses.

Data must be paired. The difference between the two groups must be normally distributed in the population or the sample size must be at least 30.

The possible combinations of null and alternative hypotheses are:

Research Question	Is the mean difference different from 0?	Is the mean difference greater than 0?	Is the mean difference less than 0?
Null Hypothesis, $H_{0}$	$\mu_d = 0$	$\mu_d = 0$	$\mu_d = 0$
Alternative Hypothesis, $H_{a}$	$\mu_d \neq 0$	$\mu_d > 0$	$\mu_d < 0$
Type of Hypothesis Test	Two-tailed, non-directional	Right-tailed, directional	Left-tailed, directional

Where $\mu_d$ is the hypothesized difference in the population.

Calculate an appropriate test statistic.

The calculation of the test statistic for dependent samples is similar to the calculation you performed earlier in this lesson for a single sample mean. In this formula, $\bar{x}_d$ is used in place of $\bar{x}$ and $s_d$ is used in place of $s$:
Test Statistic for Dependent Means

\[ t=\dfrac{\bar{x}_d-\mu_0}{\dfrac{s_d}{\sqrt{n}}} \]
- $\bar{x}_d$ = observed sample mean difference
- $\mu_0$ = mean difference specified in the null hypothesis
- $s_d$ = standard deviation of the differences
- $n$ = sample size (i.e., number of unique individuals)
Observed Sample Mean Difference

\[ \bar{x}_d=\dfrac{\Sigma{x}_d}{n} \]
- $x_d$ = observed difference
Standard Deviation of the Differences

\[ s_d=\sqrt{\dfrac{\sum (x_d-\bar{x}_d)^{2}}{n-1}} \]
Determine the p-value associated with the test statistic.

When testing hypotheses about a mean difference, a t-distribution is used to find the $p$ value. The degrees of freedom are equal to $n-1$ where $n$ is the number of pairs.
Decide between the null and alternative hypotheses.

If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis.
State a “real world” conclusion.

Based on your decision in Step 4, write a conclusion in terms of the original research question.

Example 8.33 (Quiz Scores) Below is an example of conducting a paired means t-test by hand using raw data. Next, you will learn how this can be conducted most efficiently in Minitab.

Research question: Are scores on two quizzes different?

Data were collected from 9 students and a paired means t-test was performed using hand calculations:

Student ID	Quiz 1	Quiz 2
001	98	94
002	100	98
003	95	98
004	90	88
005	90	89
006	92	91
007	80	84
008	78	80
009	88	88

Check assumptions and write hypotheses.

There are two assumptions: (1) data are paired and (2) distribution of differences is normally distribution in the population or the sample size is at least 30. The data are paired because for each student we have a quiz 1 and a quiz 2 score. We do not know if the differences are normally distributed in the population and the sample size is small, but in the video above we created a histogram of the differences and found that the sample was approximately normally distributed, so this assumption has been met, and we can perform a paired means t-test.

Given $\mu_d = \mu_1 - \mu_2$, our hypotheses are:
$H_0: \mu_d = 0$
$H_a: \mu_d \ne 0$

Calculate test statistic.

Student ID	Quiz 1	Quiz 2	Difference ($X_d$)	$X_d - \bar{X}_d$	$(X_d - \bar{X}_d)^2$
001	98	94	4	3.889	15.123
002	100	98	2	1.889	3.568
003	95	98	-3	-3.111	9.679
004	90	88	2	1.889	3.568
005	90	89	1	0.889	0.790
006	92	91	1	0.889	0.790
007	80	84	-4	-4.111	16.901
008	78	80	-2	-2.111	4.457
009	88	88	0	-0.111	0.012

Mean of the differences: $\bar{X}_d=\dfrac{\Sigma{X}_d}{n}=\dfrac{1}{9}$

For a review of computing standard deviation, see Lesson 2.

Sum of squares: $\Sigma (X_d - \bar{X}_d)^2 = 54.889$

Standard deviation of the differences: $s_d=\sqrt{\dfrac{\sum (X_d-\bar{X}_d)^{2}}{n-1}} = \sqrt{\dfrac{54.889}{9-1}}=2.619$

Test statistic: $t=\dfrac{\bar{X}_d- \mu_0}{\dfrac{s_d}{\sqrt{n}}}=\dfrac{\dfrac{1}{9}}{\dfrac{2.619}{\sqrt{9}}}=0.127$

$df=n-1=9-1=8$

Determine p-value.

We can construct a t-distribution with 8 degrees of freedom and determine what proportion of the curve falls beyond a $t$ score of 0.127. This is a two-tailed test, so we need to take into account both the left and right sides of the curve.

Fig 8.24

\[ p=0.4510+0.4510=0.9020 \]
Make a decision.

We will compare our $p$-value from step 3 to a standard alpha level of 0.05.

Because $p>\alpha$, we fail to reject the null hypothesis.
State conclusion.

There is not sufficient evidence to state that scores on the two quizzes are different.

Note!
The following video uses Minitab Express not Minitab to find the p-value

Video 8.9: Paired Means t Test Quiz Scores - Hand Calculations Example

Minitab: Paired Means Test

The steps for constructing a confidence interval or conducting a paired means t-test in Minitab are identical. The output that the procedure provides includes both the confidence interval and the $p$-value for determining statistical significance.

Conduct a paired t hypothesis test in Minitab:

Select Stat > Basic Statistics > Paired t.
Enter the appropriate columns under Sample 1 and Sample 2.
Select Perform hypothesis test.
Enter the null-hypothesis value.
Select the Options tab.
Select the appropriate alternative hypothesis.
Select OK.

Example 8.34 (Minitab: Conducting a Paired Means Test) Let’s compare students’ SAT-Math scores to their SAT-Verbal scores.

Research question: Do students score differently on the SAT-Math and SAT-Verbal tests?

Check assumptions and write hypotheses.

Because the sample size is large ($n \ge 30$), the t-distribution may be used to approximate the sampling distribution.

$H_{0}:\mu_d=0$
$H_{a}:\mu_d \ne 0$
Calculate the test statistic.

Use Minitab to conduct the paired means test:
1. Open the Minitab file: (class_survey.mpx).
2. Select Stat > Basic Statistics > Paired t.
3. Select Each sample is in a column since the data are in the worksheet.
4. Insert the variable SATM into the Sample 1 box.
5. Insert the variable SATV into the Sample 2 box.
6. Select OK.
This should result in the following output:

Paired t: SATM, SATV

Descriptive Statistics

Sample N Mean StDev SE Mean

SATM 215 599.81 84.70 5.78

SATV 215 580.33 82.44 5.62

Estimation for Paired Difference

Mean StDev SE Mean 95% CI for $\mu_d$

19.49 89.81 6.12 (7.42, 31.56)

$\mu_{\text{difference}}$: population mean of (SATM - SATV)

Test

Null hypothesis: $H_0\colon \mu_{\text{difference}} = 0$
Alternative hypothesis: $H_1\colon \mu_{\text{difference}} \ne 0$

T-Value P-Value

3.18 0.002

From the Minitab output, the t-test statistic is 3.18.
Determine the p-value associated with the test statistic.

From the output, the p-value is 0.002.
Make a decision.

$p\leq .05$; therefore, our decision is to reject the null hypothesis.
State a “real world” conclusion.

There is evidence that in the population, on average, students’ SAT-Math and their SAT-Verbal scores are different.

Sample	N	Mean	StDev	SE Mean
SATM	215	599.81	84.70	5.78
SATV	215	580.33	82.44	5.62

Mean	StDev	SE Mean	95% CI for \(\mu_d\)
19.49	89.81	6.12	(7.42, 31.56)

T-Value	P-Value
3.18	0.002

Try It! Marriage Age (Summarized Data)

In a sample of 105 married heterosexual couples, the average age difference (husband’s age - wife’s age) was 2.829 years with a standard deviation of 4.995 years. These summary statistics were taken from a data set from the Lock⁵ textbook. Is there convincing evidence that, on average, in the population, husbands tend to be older than their wives?

Check assumptions and write hypotheses.

Because the sample size is large ($n \ge 30$), the t distribution may be used to approximate the sampling distribution.

Let $\mu_d$ be the population mean difference in age (husband’s age minus wife’s age).

$H_{0}:\mu_d=0$
$H_{a}:\mu_d \gt 0$
Calculate the test statistic.

Use Minitab to conduct the paired means test using the summarized differences:
1. Select Stat > Basic Statistics > Paired t.
2. Select Summarized data (differences).
3. Enter 105 for the sample size, 2.829 for the sample mean, and 4.995 for the standard deviation.
4. Select Options.
5. Enter 0 for the hypothesized difference.
6. For the alternative hypothesis, select Difference > hypothesized difference.
7. Select OK, then OK.
This should result in the following output:

Estimation for Paired Difference

Mean StDev SE Mean 95% CI for $\mu_d$

2.829 4.995 0.487 (1.862, 3.796)

$\mu_{\text{difference}}$: population mean of (husband’s age - wife’s age)

Test

Null hypothesis: $H_0\colon \mu_{\text{difference}} = 0$
Alternative hypothesis: $H_1\colon \mu_{\text{difference}} > 0$

T-Value P-Value

5.80 0.000

From the Minitab output, the t-test statistic is 5.80.
Determine the p-value associated with the test statistic.

From the output, the p-value is reported as 0.000. We write this as $p < 0.001$.
Make a decision.

$p\leq .05$; therefore, our decision is to reject the null hypothesis.
State a “real world” conclusion.

There is convincing evidence that in the population, on average, the husband’s age in heterosexual couples is greater than the wife’s age.

Mean	StDev	SE Mean	95% CI for \(\mu_d\)
2.829	4.995	0.487	(1.862, 3.796)

T-Value	P-Value
5.80	0.000

8.4 Summary

In this lesson, you learned how to construct confidence intervals and conduct hypothesis tests for one proportion, one mean, and paired means using the z- and t-distributions. These procedures are sometimes referred to as “traditional” methods, in contrast to the bootstrapping and randomization procedures that we learned earlier in the course which are “simulation-based inference” methods.

The table below summarizes the procedures covered in this lesson. Note the similarities between all the confidence interval formulas and all of the test statistic formulas. These all follow the same general format.

Procedure	Assumptions	Standard Error	Confidence Interval	Test Statistic
One Proportion $z$	Confidence interval: At least 10 successes and 10 failures in the sampleHypothesis test: $n p_0 \ge 10$ and $n (1-p_0) \ge 10$	$\sqrt{\dfrac{p(1-p)}{n}}$	$\widehat{p} \pm z^{*} \left ( \sqrt{\dfrac{\hat{p} (1-\hat{p})}{n}} \right)$	$z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}$
One Mean $t$	Sample size is at least 30 OR population is normally distributedPopulation standard deviation is unknown*	$\dfrac{\sigma}{\sqrt n}$	$\bar{x} \pm t^{*} \dfrac{s}{\sqrt{n}}$	$t=\dfrac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}$
Paired Means $t$	Sample size is at least 30 OR differences are normally distributed in the population	$\dfrac{\sigma_d}{\sqrt n}$	$\bar{x}_d \pm t^* \left(\dfrac{s_d}{\sqrt{n}}\right)$	$t=\dfrac{\bar{x}_d-\mu_0}{\frac{s_d}{\sqrt{n}}}$

* If the population standard deviation is known, then the sampling distribution may be approximated by a z-distribution.

This lesson may be overwhelming if you are trying to memorize all the formulas that were introduced. Luckily, doing so is unnecessary. You should focus on seeing the similarities between all the procedures. For one sample mean, one sample proportion, paired means, two independent means, and two independent proportions, the following general formulas can be applied

General Form of a Confidence Interval

\[ \text{sample statistic}\pm(\text{multiplier})\ (\text{standard error}) \]

General Form of a Test Statistic

\[ \text{test statistic}=\dfrac{\text{sample statistic}-\text{null parameter}}{\text{standard error}} \]

Also, when given raw data, use Minitab. It is much more efficient than performing calculations by hand.

--- categories: [Statistical Inference, One Proportion, One Mean, Paired Means, t-Distributions] image: /assets/L8card.png # change the image path --- # Inference for One Sample ::: objectiveblock ## <i class="bi bi-check2-circle"></i>Objectives {.unnumbered .unlisted} Upon completion of this lesson, you should be able to: 1. Identify situations in which the z- or t-distributions may be used to approximate a sampling distribution. 2. Construct a confidence interval to estimate a population proportion, mean, or difference in paired means by hand given summary data. 3. Construct a confidence interval to estimate a population proportion, mean, or difference in paired means using Minitab given summary or raw data. 4. Determine the necessary minimum sample size to construct a confidence interval for a single proportion or single mean with a given level of confidence and margin of error. 5. Conduct a hypothesis test using the appropriate common distribution for a single proportion, single mean, and paired means by hand given summary data. 6. Conduct a hypothesis test using the appropriate common distribution for a single proportion, single mean, and paired means using Minitab given summary or raw data. ::: []{.bi .bi-book .fs-4} This lesson corresponds to Sections 6.1, 6.2, and 6.5 in the Lock<sup>5</sup> textbook. The general form of confidence intervals and test statistics will be the same for all the procedures covered in this lesson: ::: {.callout-note appearance="minimal"} [**General Form of a Confidence Interval**]{.fs-4} $$ \text{sample statistic}\pm(\text{multiplier})\ (\text{standard error}) $$ ::: ::: {.callout-note appearance="minimal"} [**General Form of a Test Statistic**]{.fs-4} $$ \text{test statistic}=\dfrac{\text{sample statistic}-\text{null parameter}}{\text{standard error}} $$ ::: We will be using a five-step hypothesis testing procedure: ::: {.steps} 1. **Check assumptions and write hypotheses.** The assumptions will vary depending on the test. The null and alternative hypotheses will also be written in terms of population parameters; the null hypothesis will always contain the equality (i.e., =). 2. **Calculate the test statistic.** This will vary depending on the test, but it will typically be the difference observed between the sample and population divided by a standard error. In this lesson, we will see z- and t-test statistics. Minitab will compute the test statistic. 3. **Determine the p-value.** This can be found using Minitab. 4. **Make a decision.** If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. 5. **State a "real world" conclusion.** Based on your decision in step 4, write a conclusion in terms of the original research question. ::: Some steps may vary depending on the test. We will be relying heavily on Minitab in this lesson and all the following lessons. Note that help is available in Minitab. ## One Sample Proportion One sample proportion tests and confidence intervals are covered in Section 6.1 of the Lock<sup>5</sup> textbook. In the last lesson you were introduced to the general concept of the Central Limit Theorem. The Central Limit Theorem states that if the sample size is sufficiently large then the sampling distribution will be approximately normally distributed for many frequently tested statistics, such as those that we have been working with in this course. When discussion proportions, we sometimes refer to this as the **Rule of Sample Proportions**. According to the Rule of Sample Proportions, if $np\geq 10$ and $n(1-p) \geq 10$ then the sampling distributing will be approximately normal. When constructing a confidence interval $p$ is not known but may be approximated using $\widehat p$. When conducting a hypothesis test, we check this assumption using the hypothesized proportion (i.e., the proportion in the null hypothesis). If assumptions are met, the sampling distribution will have a standard error equal to $\sqrt{\frac{p(1-p)}{n}}$. This method of constructing a sampling distribution is known as the **normal approximation method**. If the assumptions for the normal approximation method are not met (i.e., if $np$ or $n(1-p)$ is not at least 10), then the sampling distribution may be approximated using a binomial distribution. This is known as the **exact method**. This course does not cover the exact method in detail, but you will see how these tests may be performed using Minitab. ### Confidence Intervals In the following section you will see how a confidence interval for a population proportion can be constructed by hand using the normal approximation method. Using Minitab, you will learn how to construct a confidence interval for a proportion using the normal approximation method or the exact method. When given the option, it is recommended that you use Minitab as opposed to performing calculations by hand. #### Normal Approximation Formulas For the following procedures, the assumption is that both $np \geq 10$ and $n(1-p) \geq 10$. When we're constructing confidence intervals $p$ is typically unknown, in which case we use $\widehat{p}$ as an estimate of $p$. Note that $n \widehat p$ is the number of successes in the sample and $n(1- \widehat p)$ is the number of failures in the sample. This means that our sample needs to have at least 10 "successes" and at least 10 "failures" in order to construct a confidence interval using the normal approximation method. Below is the general form of a confidence interval. ::: {.callout-note appearance="minimal"} [**General Form of Confidence Interval**]{.fs-4} $$ \text{sample statistic}\pm\underbrace{(\text{multiplier})\ (\text{standard error})}_{\textbf{margin of error}} $$ ::: The sample statistic here is the sample proportion, $\widehat p$. When using the normal approximation method the multiplier is taken from the standard normal distribution (i.e., z-distribution). And, the standard error is computed using $\widehat p$ as an estimate of $p$: $$ \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} $$ This leaves us with the following formula to construct a confidence interval for a population proportion: ::: {.callout-note appearance="minimal"} [**Confidence Interval of $p$: Normal Approximation Method**]{.fs-4} $$ \underbrace{\widehat{p}}_{\text{sample statistic}} \pm \overbrace{z^{*}}^{\text{multiplier}} \underbrace{\left (\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)}_{\text{standard error}} $$ ::: ##### Finding the $z^{*}$ Multiplier {.unnumbered .unlisted} The value of the $z^*$ multiplier depends on the level of confidence. The multiplier for the confidence interval for a population proportion can be found using the standard normal distribution \[i.e., z-distribution, N(0,1)\]. The most commonly used level of confidence is 95%. As shown on the probability distribution plot below, the multiplier associated with a 95% confidence interval is 1.960, often rounded to 2 (recall the empirical rule and 95% rule). ![](assets/Multiplierfor95CI.png){#fig-Multiplierfor95CI fig-alt="Standard normal distribution showing the z multipliers for a 95% confidence interval." .mx-auto .d-block width="60%" .lightbox} Below is a table of frequently used $z^*$ multipliers. | Confidence Level | $z^*$ Multiplier | | ---------------- | :-----------------------: | | 90% | 1.645 | | 95% | 1.960, often rounded to 2 | | 98% | 2.327 | | 99% | 2.576 | : {.w-auto .table-sm .mx-auto .row-header} The value of the multiplier increases as the confidence level increases. This leads to wider intervals for higher confidence levels. We are more confident of catching the population value when we use a wider interval. ::: {#exm-pa-residency} #### PA Residency ::: {.w-75 .mx-auto #vid-8-1} {{< video https://youtu.be/YxWYT_yVa8c?si=ufdlv-aAzxTp2f-b >}} Confidence Interval for a Proportion, PA Residency Example ::: ::: ::: {#exm-dog-ownership} #### Dog Ownership In Spring 2016, a sample of 522 World Campus students were surveyed and asked if they own a dog. Of the 522 students in the sample, 273 said that they did have a dog. Construct a 95% confidence interval for the proportion of all World Campus students who have a dog. ::: {.w-75 .mx-auto #vid-8-2} {{< video https://youtu.be/VE7HCDffl2E?si=Ejcdj2P-jJB118Ht >}} 95% Confidence Interval for Proportion of Dog Owners ::: ::: ::: {#exm-books} #### Books ::: {.w-75 .mx-auto #vid-8-3} {{< video https://youtu.be/CN1RZ5QFmYQ?si=iUtpA8gPeDp658F8 >}} STAT 200 95% Confidence Interval for P Example: Paper Copy of Textbook ::: ::: :::{#exm-retirement} #### Retirement In a representative sample of 1168 American adults, 747 said they were not financially prepared for retirement. Let's construct a 95% confidence interval to estimate the proportion of all American adults who are not financially prepared for retirement. First, we need to check our assumptions that both $n\widehat p \geq 10$ and $n(1-\widehat p) \geq 10$. $$ \widehat{p}=\frac{747}{1168}=0.640 $$ $$ np=1168 (0.640) = 747 \text{ and } n(1-\widehat p)=1168(1-0.640)=421 $$ Both are greater than 10, so this assumption has been met. This means we can use the normal approximation method to construct this confidence interval. Next, we can compute the standard error. $$ \text{SE}=\sqrt{\frac{\hat{p} (1-\hat{p})}{n}}=\sqrt{\frac{0.640 (1-0.640)}{1168}}=0.014 $$ The $z^*$ multiplier for a 95% confidence interval is 1.960. The formula for a confidence interval for a proportion is $\widehat{p}\pm z^* (\text{SE})$. $$ 0.640\pm 1.960(0.014)=0.640\pm0.028=[0.612, 0.668] $$ We are 95% confident that between 61.2% and 66.8% of all American adults are not financially prepared for retirement. []{.bi .bi-question-circle-fill .fs-4 .text-success} **What if we wanted a 99% confidence interval?** Let’s think about how our interval will change. The 99% confidence interval will be wider than the 95% confidence interval. In order to increase our level of confidence, we will need to expand the interval. In terms of computing the 99% confidence interval, we will use the same point estimate $\widehat{p}$ and the same standard error. Only the multiplier will change. From the plot below, we see that the $z^*$ multiplier for a 99% confidence interval is 2.576. ![](assets/Multiplierfor99CI.png){#fig-Multiplierfor99CI fig-alt="Standard normal distribution showing the z multipliers for a 99% confidence interval" .mx-auto .d-block width="60%" .lightbox} 99\% CI: $0.640\pm 2.576 (0.014)=0.0640\pm 0.036=[0.604, 0.676]$ We are 99% confident that between 60.4% and 67.6% of all American adults are not financially prepared for retirement. ::: #### ![ ](assets/minitab.png){.icon-image} Minitab: Confidence Interval for a Proportion {{< include _minitab/construct-a-one-proportion-confidence-interval-with-raw-data.qmd >}} Before we can construct a confidence interval for a proportion, we must first determine if we should use the exact method or the normal approximation method. Recall that if $np \geq 10$ and $n(1-p) \geq 10$, then the sampling distribution can be approximated by a normal distribution. Since we don't have the population proportion ($p$), we will use $\widehat p$ as an estimate. Note that $n\widehat p$ is the number of successes in the sample and $n(1-\widehat p)$ is the number of failures in the sample. If this assumption has not been met, then the sampling distribution is constructed using a binomial distribution which Minitab refers to as the "exact method." To check this assumption, we can construct a frequency table. You first learned how to construct a frequency table in @sec-visual-representations of these online notes. Here is another example: ::: {#exm-dog-ownership-freq-table .example-minitab} #### Minitab: Frequency Tables To create a frequency table of dog ownership in Minitab: 1. Open the data set: ([fall2016stdata.mpx](data_files/fall2016stdata.mpx){download="" target="_blank"}) 2. From the toolbar in Minitab, select **Stat > Tables > Tally Individual Variables**. 3. Insert the variable *Dog* in the box on the left into the **Variable** box on the right. 4. Under Display, choose **Counts**. 5. Select OK. This should result in the following frequency table: | Dog | Count | | --- | :---: | | No | 252 | | Yes | 272 | | N= | 524 | | *= | 1 | : {.w-auto .table-sm .mx-auto .row-header} From the frequency table above we can see that there were at least 10 "successes" and at least 10 "failures" in the sample. In this example a success is defined as answering "yes" to the question "do you own a dog?" A failure is defined as answering "no." Because both $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$, the normal approximation method may be used. In Minitab, the exact method is the default method. If there are at least 10 successes and at least 10 failures, then you need to change the method to the normal approximation method. ::: ::: {#exm-dog-ownership-ci .example-minitab} #### Minitab: Confidence Interval for a Proportion (Normal Approximation) To create a 95% confidence interval of dog ownership using the normal approximation method in Minitab: 1. Open the data set: ([fall2016stdata.mpx](data_files/fall2016stdata.mpx){download="" target="_blank"}) 2. In Minitab, select **Stat > Basic Statistics > 1-Proportion**. 3. In this case we have our data in the Minitab worksheet so we will use the default **One or more samples each in a column**. 4. Insert the variable *Dog* in the box on the left into the box on the right. 5. Select **Options**. 6. The default **Confidence level** is 95. 7. Change the **Method** to **Normal approximation** because the assumption of $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$ has been met. 8. Select OK. This should result in the following output: ::: {.minitab_output} ##### Method {.unnumbered .unlisted} Event: Dog = Yes p: proportion where Dog = Yes Normal approximation is used for this analysis. ##### Descriptive Statistics | N | Event | Sample p | 95% CI for p | | --- |--- |--- |--- | | 524 | 272 | 0.519084 | (0.476304, 0.561863) | : {.w-auto .table-sm .row-header} ::: ::: ::: {.callout-caution appearance="minimal"} ##### Considerations \ []{.bi .bi-question-circle-fill .fs-4 .text-warning} **What if the assumption is not met?** If the number of successes or the number of failures in the sample is less than 10, then the exact method should be used instead of the normal approximation method. In Minitab, this means that in step 8 above the default setting of *Exact method* should not be changed. []{.bi .bi-question-circle-fill .fs-4 .text-warning} **What if we have summarized data and not data in a Minitab worksheet?** If you do not have a Minitab worksheet filled with data concerning individuals, but instead have summarized data (e.g., the number of successes and the number of failures), you would not load the data set, but in step 3 you would select **Summarized data**. For **Number of events,** enter the number of successes (i.e., $n \widehat p$) and for **Number of trials,** enter the total sample size (i.e., $n$). ::: ### <i class="bi bi-question-octagon-fill text-warning"></i> Try It! {.unnumbered .unlisted} ::: {.tryit} 1. (**Lactose Intolerance**) In a sample of 100 African American adults, 70 were identified as having some level of lactose intolerance. Use Minitab to compute a 95% confidence interval to estimate the proportion of all African American adults who have some level of lactose intolerance. ```{=html} <button id="header-1" class="btn btn-outline-success collapsed" type="submit" data-bs-toggle="collapse" data-bs-target="#panel-1" aria-expanded="false" aria-controls="panel-1"> Answer </button> <div class="collapse" id="panel-1" aria-labelledby="header-1"> <div class="card card-body"> ``` To create a 95% confidence interval of lactose intolerance using the normal approximation method in Minitab: 1. In Minitab, select **Stat > Basic Statistics > 1-Proportion**. 2. In this case we have summarized data so select **Summarized data** in the dropdown. 3. For number of events, add 70 and for number of trials add 100. 4. Select **Options**. 5. The default **Confidence level** is 95. 6. Change the **Method** to **Normal approximation** because the assumption of $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$ has been met. 7. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Method {.unnumbered .unlisted} p: event proportion Normal approximation is used for this analysis. | N | Event | Sample p | 95% CI for p | | --- |--- |--- |--- | | 100 | 70 | 0.700000 | (0.610183, 0.789817) | : Descriptive Statistics {.w-auto .table-sm .row-header} ::: ```{=html} </div> </div> ``` 2. (**Dieting**) At the beginning of the Fall 2016 semester a representative sample of World Campus STAT 200 students was surveyed. The students were asked if they were currently dieting to lose weight. In the sample of 524 students, 184 said that they were dieting to lose weight. Use Minitab to construct a 95% confidence interval for the proportion of all World Campus STAT 200 students who are dieting to lose weight. ```{=html} <button id="header-2" class="btn btn-outline-success collapsed" type="submit" data-bs-toggle="collapse" data-bs-target="#panel-2" aria-expanded="false" aria-controls="panel-2"> Answer </button> <div class="collapse" id="panel-2" aria-labelledby="header-2"> <div class="card card-body"> ``` 1. In Minitab, select **Stat > Basic Statistics > 1-Proportion**. 2. In this case we have summarized data, so select **Summarized data** in the dropdown. 3. For number of events, add 184 and for number of trials add 524. 4. Select **Options**. 5. The default **Confidence level** is 95. 6. Change the **Method** to **Normal approximation** because the assumption of $n \widehat p \geq 10$ and $n(1- \widehat p) \geq 10$ has been met. 7. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Method {.unnumbered .unlisted} p: event proportion Normal approximation is used for this analysis. ##### Descriptive Statistics | N | Event | Sample p | 95% CI for p | | --- |--- |--- |--- | | 524 | 184 | 0.351145 | (0.310276, 0.392015) | : {.w-auto .table-sm .row-header} ::: ```{=html} </div> </div> ``` ::: ### Computing Necessary Sample Size When we begin a study to estimate a population parameter we typically have an idea as how confident we want to be in our results and within what degree of accuracy. This means we get started with a set level of confidence and margin of error. We can use these pieces to determine a minimum sample size needed to produce these results by using algebra to solve for $n$: ::: {.callout-note appearance="minimal"} [**Finding Sample Size for Estimating a Population Proportion**]{.fs-4} $$ n=\left ( \dfrac{z^*}{M} \right )^2 \tilde{p}(1-\tilde{p}) $$ - $M$ is the margin of error - $\tilde p$ is an estimated value of the proportion ::: If we have no preconceived idea of the value of the population proportion, then we use $\tilde{p}=0.50$ because it is most conservative, and it will give use the largest sample size calculation. ::: {#exm-no-estimate} #### No Estimate We want to construct a 95% confidence interval for $p$ with a margin of error equal to 4%. Because there is no estimate of the proportion given, we use $\tilde{p}=0.50$ for a conservative estimate. For a 95% confidence interval, $z^*=1.960$. $$ n=\left ( \dfrac{1.960}{0.04} \right )^2 (0.5)(1-0.5)=600.25 $$ This is the minimum sample size, therefore we should round up to 601. In order to construct a 95% confidence interval with a margin of error of 4%, we should obtain a sample of at least $n=601$. ::: ::: {#exm-estimate-known} #### Estimate Known We want to construct a 95% confidence interval for $p$ with a margin of error equal to 4%. What if we knew that the population proportion was around 0.25? The $z^*$ multiplier for a 95% confidence interval is 1.960. Now, we have an estimate to include in the formula: $$ n=\left ( \dfrac{1.960}{0.04} \right )^2 (0.25)(1-0.25)=450.188 $$ Again, we should round up to 451. In order to construct a 95% confidence interval with a margin of error of 4%, given $\tilde{p}=.25$, we should obtain a sample of at least $n=451$. Note that when we changed $\tilde{p}$ in the formula from .50 to .25, the necessary sample size decreased from $n=601$ to $n=451$. ::: ### Hypothesis Testing A hypothesis test for a proportion is used when you are comparing one group to a known or hypothesized population proportion value. In other words, you have one sample with one categorical variable. The hypothesized value of the population proportion is symbolized by $p_0$ because this is the value in the null hypothesis ($H_0$). If $np_0 \ge 10$ and $n(1-p_0) \ge 10$ then the distribution of sample proportions is approximately normal and can be estimated using the normal distribution. That sampling distribution will have a mean of $p_0$ and a standard deviation (i.e., standard error) of $\sqrt{\frac{p_0 (1-p_0)}{n}}$. Recall that the standard normal distribution is also known as the z-distribution. Thus, this is known as a "single sample proportion z-test" or "one sample proportion z-test." If $np_0 < 10$ or $n(1-p_0) < 10$ then the distribution of sample proportions follows a binomial distribution. We will not be conducting this test by hand in this course, however you will learn how this can be conducted using Minitab using the exact method. #### Normal Approximation Method Formulas Here we will be using the five-step hypothesis testing procedure to compare the proportion in one random sample to a specified population proportion using the normal approximation method. ::: {.steps} 1. **Check assumptions and write hypotheses.** In order to use the normal approximation method, the assumption is that both $n p_0 \geq 10$ and $n (1-p_0) \geq 10$. Recall that $p_0$ is the population proportion in the null hypothesis. | Research Question | Is the proportion different from $p_0$? | Is the proportion greater than $p_0$? | Is the proportion less than $p_0$? | | --- |--- |--- |--- | | Null Hypothesis, $H_{0}$ | $p=p_0$ | $p= p_0$ | $p= p_0$ | | Alternative Hypothesis, $H_{a}$ | $p\neq p_0$ | $p> p_0$ | $p< p_0$ | | Type of Hypothesis Test | Two-tailed, non-directional | Right-tailed, directional | Left-tailed, directional | : {.w-auto .table-sm .mx-auto .row-header} Where $p_0$ is the hypothesized population proportion that you are comparing your sample to. 2. **Calculate the test statistic.** When using the normal approximation method we will be using a z-test statistic. The z-test statistic tells us how far our sample proportion is from the hypothesized population proportion in standard error units. Note that this formula follows the basic structure of a test statistic that you learned in the last lesson: $$ \text{test statistic}=\dfrac{\text{sample statistic}-\text{null parameter}}{\text{standard error}} $$ ::: {.callout-note appearance="minimal"} [**Test statistic: One Group Proportion**]{.fs-4} $$ z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}} $$ {#eq-test-statistic-one-group-proportion} $\widehat{p}$ = sample proportion $p_{0}$ = hypothesize population proportion $n$ = sample size ::: 3. **Determine the p-value.** Given that the null hypothesis is true, the **p-value** is the probability that a randomly selected sample of *n* would have a sample proportion as different, or more different, than the one in our sample, in the direction of the alternative hypothesis. We can find the p-value by mapping the test statistic from step 2 onto the z-distribution. ::: {.callout-caution appearance="minimal"} **Note!** \ **Note that p-values are also symbolized by $p$. Do not confuse this with the population proportion which shares the same symbol.** ::: We can look up the $p$-value using Minitab by constructing the sampling distribution. Because we are using the normal approximation here, we have a z-test statistic that we can map onto the z-distribution. Recall, the z-distribution is a normal distribution with a mean of 0 and standard deviation of 1. If we are conducting a one-tailed (i.e., right- or left-tailed) test, we look up the area of the sampling distribution that is beyond our test statistic. If we are conducting a two-tailed (i.e., non-directional) test there is one additional step: we need to multiply the area by two to take into account the possibility of being in the right or left tail. 4. **Make a decision.** We can decide between the null and alternative hypotheses by examining our p-value. If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. Unless stated otherwise, assume that $\alpha=.05$. When we reject the null hypothesis our results are said to be statistically significant. 5. **State a "real world" conclusion.** Based on our decision in step 4, we will write a sentence or two concerning our decision in relation to the original research question. ::: ::: {#exm-male-babies} #### 1 Sample Proportion Normal Approximation Method - Male Babies ::: {.w-75 .mx-auto #vid-8-4} {{< video https://youtu.be/5G1QdAYtFAM?si=-kDjBo5m85gkyBIr >}} 1 Sample Proportion Normal Approximation Method - Male Babies Example ::: ::: ::: {#exm-handedness} ##### Handedness **Research Question**: Are more than 80% of American's right-handed? In a sample of 100 Americans, 87 were right-handed. ::: {.steps} 1. **Check assumptions and write hypotheses.** $np_0 = 100(0.80)=80$ $n(1-p_0) = 100 (1-0.80) = 20$ Both $np_0$ and $n(1-p_0)$ are at least 10 so we can use the normal approximation method. This is a right-tailed test because we want to know if the proportion is greater than 0.80. $H_{0}\colon p=0.80$ $H_{a}\colon p>0.80$ 2. **Calculate the test statistic.** (@eq-test-statistic-one-group-proportion) $$\widehat{p}=\dfrac{87}{100}=0.87,\ p_{0}=0.80,\ n=100$$ $$ z= \dfrac{\widehat{p}- p_0 }{\sqrt{\dfrac{p_0 (1- p_0)}{n}}}= \dfrac{0.87-0.80}{\sqrt{\dfrac{0.80 (1-0.80)}{100}}}=1.75 $$ Our z-test statistic is 1.75. 3. **Determine the p-value associated with the test statistic.** This is a right-tailed test so we need to find the area to the right of the test statistic, $z=1.75$, on the z-distribution. Using Minitab, we find the probability $P(z\geq1.75)=0.0400592$ which may be rounded to $\text{p-value}=0.0401$. ![](assets/zgeq175.png){#fig-zgeq175 fig-alt="Distribution plot of Density vs X - Normal, Mean=0, StDev=1" .mx-auto .d-block width="60%" .lightbox} 4. **Make a decision.** $p\leq .05$, therefore our decision is to reject the null hypothesis. 5. **State a "real world" conclusion.** Yes, there is convincing evidence to state that more than 80% of all Americans are right-handed. ::: ::: ::: {#exm-ice-cream} #### Ice Cream **Research Question**: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%? In a sample of 50 customers, 60% preferred chocolate over vanilla. ::: {.steps} 1. **Check assumptions and write hypotheses.** $np_0 = 50(0.80) = 40$ $n(1-p_0)=50(1-0.80) = 10$ Both $np_0$ and $n(1-p_0)$ are at least 10. We can use the normal approximation method. This is a left-tailed test because we want to know if the proportion is less than 0.80. $H_{0}\colon p=0.80$ $H_{a}\colon p<0.80$ 2. **Calculate the test statistic.** (@eq-test-statistic-one-group-proportion) $\widehat{p}=0.60$, $p_{0}=0.80$, $n=50$ $$ z= \dfrac{\widehat{p}- p_0 }{\sqrt{\dfrac{p_0 (1- p_0)}{n}}}= \dfrac{0.60-0.80}{\sqrt{\dfrac{0.80 (1-0.80)}{50}}}=-3.536 $$ Our z-test statistic is -3.536. 3. **Determine the p-value.** This is a left-tailed test so we need to find the area to the right of our test statistic, $z=-3.536$. ![](assets/zleqneg3536.png){#fig-zleqneg3536 fig-alt="Distribution Plot of Density vs X - Normal, Mean=0, StDev=1" .mx-auto .d-block width="60%" .lightbox} From the Minitab output above, the p-value is 0.0002031. 4. **Make a decision.** $p \leq.05$, therefore our decision is to reject the null hypothesis. 5. **State a "real world" conclusion.** Yes, there is convincing evidence that the percentage of all Creamery customers who prefer chocolate ice cream over vanilla is less than 80%. ::: ::: ::: {#exm-overweight-citizens} #### Overweight Citizens According to the Center for Disease Control (CDC), the percent of adults 20 years of age and over in the United States who are overweight is 69.0% (see [NCHS FastStats: Obesity and Overweight](http://www.cdc.gov/nchs/fastats/obesity-overweight.htm)). One city’s council wants to know if the proportion of overweight citizens in their city is different from this known national proportion. They take a random sample of 150 adults 20 years of age or older in their city and find that 98 are classified as overweight. Let’s use the five-step hypothesis testing procedure to determine if there is convincing evidence that the proportion in this city is different from the known national proportion. ::: {.steps} 1. **Check assumptions and write hypotheses.** $np_0 =150 (0.690)=103.5$ $n (1-p_0) =150 (1-0.690)=46.5$ Both $n p_0$ and $n (1-p_0)$ are at least 10, this assumption has been met. **Research question:** Is this city’s proportion of overweight individuals different from 0.690? This is a non-directional test because our question states that we are looking for a difference as opposed to a specific direction. This will be a two-tailed test. $H_{0}\colon p=0.690$ $H_{a}\colon p\neq 0.690$ 2. **Calculate the test statistic.** (@eq-test-statistic-one-group-proportion) $$ \widehat{p}=\dfrac{98}{150}=.653 $$ $$ z =\dfrac{0.653- 0.690 }{\sqrt{\dfrac{0.690 (1- 0.690)}{150}}} = -0.980 $$ Our test statistic is $z=-0.980$. 3. **Determine the p-value.** This is a non-directional (i.e., two-tailed) test, so we need to find the area under the z-distribution that is more extreme than $z=-0.980$. In Minitab, we find the proportion of a normal curve beyond $\pm0.980$: ![](assets/z2tails98.png){#fig-z2tails98 fig-alt="Distribution Plot of Density vs X - Normal, Mean=0, StDev=1" .mx-auto .d-block width="60%" .lightbox} $$ \text{p-value}=0.163543+0.163543=0.327086 $$ 4. **Make a decision.** $p>\alpha$, therefore we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not sufficient evidence to state that the proportion of citizens of this city who are overweight is different from the national proportion of 0.690. ::: ::: #### ![ ](assets/minitab.png){.icon-image} Minitab: Hypothesis Tests for One Proportion A hypothesis test for one proportion can be conducted in Minitab. This can be done using raw data or summarized data. - If you have a data file with every individual's observation, then you have **raw data**. - If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data. The next two examples will show you how to use Minitab to conduct this analysis using either **raw data** or **summarized data**. Note that the default method for constructing the sampling distribution in Minitab is to use the exact method. If $np_0 \geq 10$ and $n(1-p_0) \geq 10$ then you will need to change this to the normal approximation method. This must be done manually. **Minitab will use the method that you select, it will not check assumptions for you!** ##### ![ ](assets/minitab.png){.icon-image} Minitab: 1 Proportion z-Test, Raw Data If you have data in a Minitab worksheet, then you have what we call "raw data." This is in contrast to "summarized data" which you'll see in the next section. In order to use the normal approximation method both $np_0 \geq 10$ and $n(1-p_0) \geq 10$. Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. This must be checked manually. **Minitab will not check assumptions for you.** {{< include _minitab/conduct-a-one-proportion-hypothesis-test-with-raw-data.qmd >}} ::: {#exm-male-st .example-minitab} #### Minitab: Conducting a One Sample Proportion z-Test: Raw Data The data for this example come from the class survey, which includes a sample of 226 students. **Research question**: Is the proportion of students who are male different from 0.50? ::: {.steps} 1. **Check assumptions and write hypotheses.** Here, $n=226$ and $p_0=0.50$. The expected numbers of successes and failures under the null hypothesis are $$np_0=226(0.50)=113$$ and $$n(1-p_0)=226(1-0.50)=113.$$ Both expected counts are at least 10. Therefore, the assumptions are satisfied, and we can use the normal approximation method. $H_0\colon p = 0.50$ $H_a\colon p \ne 0.50$ 2. **Calculate the test statistic.** Use Minitab to conduct the one-sample proportion z-test with raw data: 1. Open the Minitab file: ([class_survey.mpx](data_files/class_survey.mpx){download="" target="_blank"}). 2. Select **Stat > Basic Statistics > 1 Proportion**. 3. Select **One or more samples, each in a column** from the dropdown. 4. Double-click the variable *Biological Sex* to insert it into the box. 5. Check **Perform hypothesis test** and enter *0.50* in the **Hypothesized proportion** box. 6. Select **Options**. 7. Use the default alternative hypothesis, **Proportion ≠ hypothesized proportion**. 8. Use the default confidence level of 95. 9. Select **Normal approximation method**. 10. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Method {.unnumbered .unlisted} Event: Biological Sex = Male p: proportion where Biological Sex = Male Normal approximation is used for this analysis. ##### Descriptive Statistics {.unnumbered .unlisted} | N | Event | Sample p | 95% CI for p | | --- |--- |--- |--- | | 226 | 99 | 0.438053 | (0.373368, 0.502738) | : {.w-auto .table-sm .row-header} ##### Test {.unnumbered .unlisted} | Null hypothesis \ H~0~: p = 0.5 | Alternative hypothesis H~1~: p ≠ 0.5 | Z-Value | P-Value | | --- |--- | | -1.86 | 0.063 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, $z = -1.86$. 3. **Determine the p-value.** From the Minitab output, $p = 0.063$. 4. **Make a decision.** $p > \alpha$, fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence that the proportion of all students in the population who are male is different from 0.50. ::: ::: ##### ![ ](assets/minitab.png){.icon-image} Minitab: 1 Proportion z-Test, Summarized Data Use the summarized-data procedure when the individual observations are not available but the number of events and total number of trials are known. As with raw data, check the normal approximation assumptions before selecting the analysis method in Minitab. {{< include _minitab/conduct-a-one-proportion-hypothesis-test-with-summarized-data.qmd >}} ::: {#exm-overweight .example-minitab} #### Minitab: Conducting a One Sample Proportion z-Test: Summarized Data The following example uses a scenario in which we want to know if the proportion of college women who think they are overweight is less than 40%. We collect data from a random sample of 129 college women and 37 said that they think they are overweight. **Research question**: Is the proportion of college women who think they are overweight less than 0.40? ::: {.steps} 1. **Check assumptions and write hypotheses.** Here, $n=129$ and $p_0=0.40$. The expected numbers of successes and failures under the null hypothesis are $$np_0=129(0.40)=51.6$$ and $$n(1-p_0)=129(1-0.40)=77.4.$$ Both expected counts are at least 10. Therefore, the assumptions are satisfied, and we can use the normal approximation method. $H_0\colon p = 0.40$ $H_a\colon p < 0.40$ 2. **Calculate the test statistic.** Use Minitab to conduct the one-sample proportion z-test with summarized data: 1. Select **Stat > Basic Statistics > 1 Proportion**. 2. Select **Summarized data** from the dropdown. 3. Enter 37 for the number of events and 129 for the number of trials. 4. Check **Perform hypothesis test** and enter *0.40* in the **Hypothesized proportion** box. 5. Select **Options**. 6. Set the alternative hypothesis to **Proportion < hypothesized proportion**. 7. Use the default confidence level of 95. 8. Select **Normal approximation method**. 9. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Method {.unnumbered .unlisted} Event: Event proportion Normal approximation is used for this analysis. ##### Descriptive Statistics {.unnumbered .unlisted} | N | Event | Sample p | 95% Upper Bound for p | | --- |--- |--- |--- | | 129 | 37 | 0.286822 | 0.352321 | : {.w-auto .table-sm .row-header} ##### Test {.unnumbered .unlisted} | Null hypothesis \ H~0~: p = 0.4 | Alternative hypothesis H~1~: p < 0.4 | Z-Value | P-Value | | --- |--- | | -2.62 | 0.004 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, $z=-2.62$. 3. **Determine the p-value.** From the Minitab output, $p=0.004$. 4. **Make a decision.** $p \leq \alpha$, reject the null hypothesis. 5. **State a "real world" conclusion.** There is convincing evidence that the proportion of women in the population who think they are overweight is less than 40%. ::: ::: ###### :::{#exm-gym-membership .example-minitab} ###### Minitab: Normal Approx. Method (Gym membership) **Research question:** Are less than 50% of all individuals with a membership at one gym female? A simple random sample of 60 individuals with a membership at one gym was collected. Each individual's biological sex was recorded. There were 24 females. ::: {.steps} 1. **Check assumptions and write hypotheses.** $np_0=60(0.50)=30$ and $n(1-p_0)=60(1-0.50)=30$. Both values are at least 10, so we can use the normal approximation method. $H_0\colon p = 0.50$ $H_a\colon p < 0.50$ 2. **Calculate the test statistic.** Use Minitab to conduct a one-sample proportion *z*-test with summarized data: 1. Select **Stat > Basic Statistics > 1 Proportion**. 2. Select **Summarized data** from the dropdown. 3. Enter **24** for the number of events and **60** for the number of trials. 4. Select **Perform hypothesis test** and enter **0.50** for the hypothesized proportion. 5. Select **Options**. 6. For the alternative hypothesis, select **Proportion < hypothesized proportion**. 7. Use the default confidence level of 95. 8. Select **Normal approximation method**. 9. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Method {.unnumbered .unlisted} Event: Event proportion Normal approximation is used for this analysis. ##### Descriptive Statistics {.unnumbered .unlisted} | N | Event | Sample p | 95% Upper Bound for p | | --- |--- |--- |--- | | 60 | 24 | 0.400000 | 0.504030 | : {.w-auto .table-sm .row-header} ##### Test {.unnumbered .unlisted} | Null hypothesis | $H_0: p=0.50$ | | --- |--- | | Alternative hypothesis | $H_1: p<0.50$ | : {.w-auto .table-sm .row-header} | Z-Value | P-Value | | --- |--- | | -1.55 | 0.061 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, the test statistic is $z=-1.55$. 3. **Determine the p-value.** From the output, $p=0.061$. 4. **Make a decision.** Because $p \geq \alpha$, we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not enough evidence to conclude that less than 50% of all members at this gym are female. ::: ::: ## One Sample Mean One sample mean tests are covered in Section 6.2 of the Lock<sup>5</sup> textbook. Concerning one sample mean, the **Central Limit Theorem** states that if the sample size is large, then the distribution of sample means will be approximately normally distributed with a standard deviation (i.e., standard error) equal to $\frac{\sigma}{\sqrt n}$. In this course, a "large" sample size will be defined as one where $n \ge 30$. When constructing confidence interval and conducting hypothesis tests we often do not know the value of $\sigma$. In those cases, $\sigma$ may be estimated using the sample standard deviation ($s$). When we are using $s$ to estimate $\sigma$ our sampling distribution will not follow a z-distribution exactly. Instead, we use what is known as the t-distribution. Like the z-distribution, the t-distribution is symmetrical. The difference is that its height varies depending on the sample size. By doing so, the distribution becomes more conservative for smaller sample sizes to account for some error that may occur from estimating $\sigma$ with $s$ from a small sample. As $n$ approaches infinity ($\infty$) the t-distribution approaches the standard normal distribution. The next section compares the z- and t-distributions. When constructing confidence intervals and conducting hypothesis tests we will usually be using the t-distribution when working with one mean. The only exception would be in cases where $\sigma$ is known. This scenario is most common in the fields of education and psychology where some tests are normed to have a certain $\mu$ and $\sigma$. In those cases, the z-distribution can be used. In terms of language, all of these tests could be called "single sample mean tests" or "one sample mean tests." We could also specify the sampling distribution by using the term "single sample mean t-test" or "single sample mean z-test." The flow chart below may help you in determining which method should be used when constructing a sampling distribution for one sample mean. Identify when z- and t-distributions should be used. - Is the population known to be normally distributed? - Is the population standard deviation known? - Is the sample size at least 30? ![Flow Chart: Approximating the sample distribution](/assets/one-sample-mean-flowchart.png){#fig-one-sample-mean-flowchart.png1 fig-alt="Flowchart of z- and t- Distributions. Description below." fig-align="center" width="80%" .lightbox} <details> <summary>Text alternative of @fig-one-sample-mean-flowchart.png1</summary> - Is the population known to be normally distributed? - Yes: - Is the population standard deviation known? - Yes: z-distribution. - No: t-distribution. - No: - Is the sample size at least 30? - Yes: t-distribution. - No: Bootstrap/Randomization. </details> ### t-Distribution The height of the t-distribution is determined by the number of degrees of freedom (df). For a one sample mean test, $df=n-1$. The first plot below compares the standard normal distribution (i.e., z-distribution) to a t-distribution. The solid blue line is the standard normal distribution and the dashed red line is a t-distribution with 2 degrees of freedom. Here, the tails of the t-distribution are higher than the tails of the normal distribution. ![](assets/Plot1.png){#fig-Plot1 fig-alt="A plot showing the z-distribution compared to a t distribution with df=2." .mx-auto .d-block width="60%" .lightbox} If you think about the area under the curve, the higher tails mean that more area will fall in the tails. For example, as seen in the following two plots, $P(z>2.00)=0.0227501$ while $P(t_{df=2}>2.00)=0.0917517$. ![](assets/Plot2.png){#fig-Plot2 fig-alt="Standard normal (i.e., z) distribution showing the area above z=2" .mx-auto .d-block width="60%" .lightbox} ![](assets/Plot3b.png){#fig-Plot3b fig-alt="Probability distribution plot showing the area greater than t=2 on a distribution with 2 degrees of freedom" .mx-auto .d-block width="60%" .lightbox} The next plot compares the standard normal distribution to a t distribution with 10 degrees of freedom. Notice that the two distributions are becoming more similar as the sample size increases. ![](assets/Plot4.png){#fig-Plot4 fig-alt="Plot comparing the z-distribution to a t distribution with 10 degrees of freedom" .mx-auto .d-block width="60%" .lightbox} The next plot compares the standard normal distribution to a t distribution with 30 degrees of freedom. ![](assets/Plot5.png){#fig-Plot5 fig-alt="Plot comparing the standard normal distribution to a t distribution with 30 degrees of freedom" .mx-auto .d-block width="60%" .lightbox} In the final graph, the standard normal distribution is compared to a t distribution with 500 degrees of freedom. Here, the two distributions are nearly identical. As the degrees of freedom approach infinity, the t distribution approaches (i.e., becomes more similar to) the standard normal distribution. ![](assets/Plot6.png){#fig-Plot6 fig-alt="Plot comparing the standard normal distribution to a t distribution with 500 degrees of freedom" .mx-auto .d-block width="60%" .lightbox} #### ![ ](assets/minitab.png){.icon-image} Minitab: Using *t*-Distribution Probability Plots The procedures for constructing t distributions in Minitab are similar to those for constructing z-distributions. We can construct a probability distribution plot to find the $t^{*}$ multiplier when constructing a confidence interval. And, we can construct a plot to find the p-value when conducting a hypothesis test. {{< include _minitab/find-a-t-star-multiplier.qmd >}} {{< include _minitab/find-a-p-value-from-a-t-test-statistic.qmd >}} ### Confidence Intervals Confidence intervals are used to estimate unknown population parameters. Because the population standard deviation ($\sigma$) will almost always be unknown in situations in which we are constructing confidence intervals for means, the t-distribution is used to estimate the sampling distribution. The following section will show you how to construct a confidence interval for a population mean using formulas and using Minitab. Similar to how we computed necessary minimum sample sizes for confidence intervals for proportions, we will also compute the necessary minimum sample size for constructing a confidence interval for a mean. #### Formulas Earlier in this lesson we considered confidence intervals for proportions and the multiplier in our intervals was a value from the standard normal (i.e., $z$) distribution. But, what if our variable of interest is a quantitative variable, and we want to estimate a population mean? We apply similar techniques when constructing a confidence interval for a mean, but now we are interested in estimating the population mean ($\mu$) by using the sample statistic ($\bar{x}$) and the multiplier is a $t$ value. Similar to the $z$ values that you used as the multiplier for constructing confidence intervals for population proportions, here you will use $t$ values as the multipliers. Because $t$ values vary depending on the number of degrees of freedom (df), you will need to use statistical software to look up the appropriate $t$ value for each confidence interval that you construct. The degrees of freedom will be based on the sample size. Since we are working with one sample here, $df=n-1$. ::: {#exm-t-multiplier .example-minitab} #### Minitab: Finding $t^*$ Multipliers To find the $t^{*}$ multiplier for a 98% confidence interval with 15 degrees of freedom: 1. In Minitab, select **Graph > Probability Distribution Plot > View Probability**. 2. Change the **Distribution** to *t*. 3. Enter 15 for the **Degrees of freedom**. 4. Select **Options**. 5. Choose **A specified probability**. 6. Select **Equal tails**. 7. For **Probability**, enter 0.02 (if there is 0.98 in the middle, then 0.02 is split equally between the left and right tails). This should result in an output similar to the output below. Note that your results may be slightly different due to random sampling variation. ![](assets/200tdistplot.png){#fig-200tdistplot fig-alt="T Distribution, 98% CI, DF=15" .mx-auto .d-block width="60%" .lightbox} ::: Let’s review some of symbols and equations that we learned in previous lessons: ::: {.formula-definitions .ms-3} Sample size : $n$ Population mean : $\mu = \dfrac{\sum X}{N}$ Sample mean : $\bar{x} = \dfrac{\sum x}{n}$ Standard error of the mean : $\operatorname{SE} = \dfrac{s}{\sqrt{n}}$ Multiplier : $t^*$ Degrees of freedom (one group) : $df = n - 1$ ::: Recall the general form for a confidence interval: ::: {.callout-note appearance="minimal"} [**General Form of Confidence Interval**]{.fs-4} $$ \text{sample statistic}\pm\underbrace{(\text{multiplier})\ (\text{standard error})}_{\textbf{margin of error}} $$ ::: When constructing a confidence interval for a population mean the point estimate is the sample mean, $\bar{x}$. The multiplier is taken from a t-distribution. And, the standard error is equal to $\frac{s}{\sqrt{n}}$. ::: {.callout-note appearance="minimal"} [**Confidence Interval for a Population Mean**]{.fs-4} $$ \underbrace{\bar{x}}_{\text{sample statistic}} \pm \overbrace{t^{*}}^{\text{multiplier}} \underbrace{ \dfrac{s}{\sqrt{n}}}_{\text{standard error}} $$ ::: The following examples walk through constructing confidence intervals for population means by hand. Then, you will learn how to compute confidence intervals using Minitab. ::: {#exm-MLB-age} #### MLB Age In a sample of 30 current MLB pitchers, the mean age was 28 years with a standard deviation of 4.4 years. Construct a 95% confidence interval to estimate the mean age of all current MLB pitchers. This is what we know: $n=30$, $\bar{x}=28$, and $s=4.4$. In order to compute the confidence interval for $\mu$ we will need the t multiplier and the standard error ($\frac{s}{\sqrt{n}}$). $$ df=n-1=30-1=29 $$ ![](assets/200tdistrmlbex.png){#fig-200tdistrmlbex fig-alt="t Distribution showing the multipliers for a 95% confidence interval given 21 degrees of freedom" .mx-auto .d-block width="60%" .lightbox} For a 95% confidence interval with 29 degrees of freedom, $t^{*}=2.045$ $$ \text{SE}=\dfrac{s}{\sqrt{n}}=\dfrac{4.4}{\sqrt{30}}=0.803 $$ Thus, our confidence interval for $\mu$ is: $$ 28\pm 2.045(0.803)=28\pm1.643=[26.357,29.643] $$ We are 95% confident that the population mean age is between 26.357 and 29.643. ::: ::: {#exm-sleep-deprivation} #### Sleep Deprivation In a class survey, students were asked how many hours they sleep per night. In the sample of 22 students, the mean was 5.77 hours with a standard deviation of 1.572 hours. That distribution was approximately normal. Let’s construct a 95% confidence interval for the mean number of hours slept per night in the population from which this sample was drawn. This is what we know: $n=22$, $\bar{x}=5.77$, and $s=1.572$. In order to compute the confidence interval for $\mu$ we will need the t multiplier and the standard error ($\frac{s}{\sqrt{n}}$). $$ df=n-1=22-1=21 $$ ![](assets/tMultdf2195.png){#fig-tMultdf2195 fig-alt="t Distribution showing the multipliers for a 95% confidence interval given 21 degrees of freedom" .mx-auto .d-block width="60%" .lightbox} For a 95% confidence interval with 21 degrees of freedom, $t^{*}=2.080$ $$ \text{SE}=\frac{s}{\sqrt{n}}=\frac{1.572}{\sqrt{22}}=0.335 $$ Thus, our confidence interval for $\mu$ is: $$ 5.77\pm 2.080(0.335)=5.77\pm0.697=[5.073, 6.467] $$ We are 95% confident that the population mean is between 5.073 and 6.467 hours. []{.bi .bi-question-circle-fill .fs-4 .text-success} **What if we wanted to be more conservative and construct a 99% confidence interval?** The only thing that would change is our multiplier. Now, $t^{*}=2.831$. ![](assets/tMultdf2199.png){#fig-tMultdf2199 fig-alt="t Distribution showing the multipliers for a 99% confidence interval given 21 degrees of freedom" .mx-auto .d-block width="60%" .lightbox} $$ 5.77\pm 2.831(0.335)=5.77\pm0.948=[4.822, 6.718] $$ We are 99% confident that the population mean is between 4.822 and 6.718 hours. ::: ::: {#exm-milk} #### Milk A study of 66,831 dairy cows found that the mean milk yield was 12.5 kg per milking with a standard deviation of 4.3 kg per milking (data from Berry, et al., 2013). Construct a 95% confidence interval for the average milk yield in the population. First, let's compute the standard error: $$ \text{SE}=\dfrac{s}{\sqrt{n}}=\dfrac{4.3}{\sqrt{66831}}=0.0166 $$ The standard error is small because the sample size is very large. Next, let's find the $t^*$ multiplier: $$ df=66831-1=66830 $$ ![](assets/tDistdf66830.png){#fig-tDistdf66830 fig-alt="Distribution Plot - T, DF=66830" .mx-auto .d-block width="60%" .lightbox} $$ t^{*}=1.960 $$ Now, we can construct our 95% confidence interval: 95% CI: $12.5\pm1.960(0.017)=12.5\pm0.033=[12.467,\;12.533]$ We are 95% confident that the mean milk yield in the population is between 12.467 and 12.533 kg per milking. ::: ::: {#exm-milk .example-minitab} #### Minitab: Confidence Interval of a Mean Here you will learn how to use Minitab to construct a confidence interval for a mean. The procedure is similar to the one that you learned earlier in this lesson for constructing a confidence interval for a proportion. The following example walks through this procedure when data are in a Minitab work. To create a 95% confidence interval of mean height in Minitab: 1. Open the data set: ([fall2016stdata.csv](data_files/fall2016stdata.csv){download="" target="_blank"}). 2. In Minitab, select **Stat > Basic Statistics > 1-sample t**. 3. In this case, we have our data in the Minitab worksheet, so we will use the default **One or more samples, each in a column**. 4. Insert the variable *Height* in the box on the left into the box on the right. 5. Select **Options**. 6. The default **Confidence level** is *95*. 7. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Descriptive Statistics | N | Mean | StDev | SE Mean | 95% CI for $\mu$ | | --- |--- |--- |--- |--- | | 525 | 67.009 | 4.462 | 0.195 | (66.627, 67.392) | : {.w-auto .table-sm .row-header} $\mu$: *mean of Height* ::: ::: {.callout-caution appearance="minimal"} []{.bi .bi-question-circle-fill .fs-4 .text-warning} **What if we have summarized data and not data in a Minitab worksheet?** \ If you do not have a Minitab worksheet filled with data concerning individuals, but instead have summarized data (e.g., the values of $s$, $\bar{x}$, and $n$), you would skip step 1 above and in step 3 you would select *Summarized data*. ::: ::: ### <i class="bi bi-question-octagon-fill text-warning"></i> Try It! {.unnumbered .unlisted} ::: {.tryit} 1. **Estimating the average MLB Pitcher's age** In a sample of 30 current MLB pitchers, the mean age was 28 years with a standard deviation of 4.4 years. Construct a 95% confidence interval to estimate the mean age of all current MLB pitchers. ```{=html} <button id="header-3" class="btn btn-outline-success collapsed" type="submit" data-bs-toggle="collapse" data-bs-target="#panel-3" aria-expanded="false" aria-controls="panel-3"> Answer </button> <div class="collapse" id="panel-3" aria-labelledby="header-3"> <div class="card card-body"> ``` We know that n = 30, $\bar{x}=28$, and s = 4.4. To create a 95% confidence interval of mean age in Minitab: 1. In Minitab, select **Stat > Basic Statistics > 1-sample t**. 2. In this case we have summarized data so select **Summarized Data** from the dropdown 3. Enter 30 for the sample size, 28 for the sample mean, and 4.4 for the standard deviation. 4. Select **Options**. 5. The default **Confidence level** is *95*. 6. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Descriptive Statistics | N | Mean | StDev | SE Mean | 95% CI for $\mu$ | | --- |--- |--- |--- |--- | | 30 | 28.000 | 4.400 | 0.803 | (26.357, 29.643) | : {.w-auto .table-sm .row-header} $\mu$: *population mean of sample* ::: We are 95% confident that the population mean age is between 26.357 and 29.643 years. ```{=html} </div> </div> ``` 2. **Coffee Sales (Data in Column)** For 48 days data concerning sales were collected from one student-run cafe. Construct a 95% confidence interval for the mean number of coffees sold per day. ```{=html} <button id="header-3" class="btn btn-outline-success collapsed" type="submit" data-bs-toggle="collapse" data-bs-target="#panel-3b" aria-expanded="false" aria-controls="panel-3b"> Answer </button> <div class="collapse" id="panel-3b" aria-labelledby="header-3"> <div class="card card-body"> ``` To create a 95% confidence interval of mean number of coffees sold per day in Minitab: 1. Open the file: ([cafedata.mpx](data_files/cafedata.mpx){download="" target="_blank"}). 2. In Minitab, select **Stat > Basic Statistics > 1-sample t**. 3. In this case the data is in a worksheet, so use **One or more samples, each in a column**. 4. Select the variable *Coffees*. 5. Select **Options**. 6. The default **Confidence level** is *95*. 7. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Descriptive Statistics | N | Mean | StDev | SE Mean | 95% CI for $\mu$ | | --- |--- |--- |--- |--- | | 47 | 21.51 | 11.08 | 1.62 | (18.26, 24.76) | : {.w-auto .table-sm .row-header} $\mu$: *population mean of Coffees* ::: We are 95% confident that the population mean number of coffees sold per day is between 18.26 and 24.76. ```{=html} </div> </div> ``` ::: ### Computing Necessary Sample Size Calculating the sample size necessary for estimating a population mean with a given margin of error and level of confidence is similar to that for estimating a population proportion. However, since the t-distribution is not as “neat” as the standard normal distribution, the process can be iterative. (Recall, the shape of the t-distribution is different for each degree of freedom). This means that we would solve, reset, solve, reset, etc. until we reached a conclusion. Yet, we can avoid this iterative process if we employ an approximate method based on t-distribution approaching the standard normal distribution as the sample size increases. This approximate method invokes the following formula: ::: {.callout-note appearance="minimal"} [**Finding the Sample Size for Estimating a Population Mean**]{.fs-4} $$ n=\dfrac{z^{2}\widetilde{\sigma}^{2}}{M^{2}}=\left ( \dfrac{z\widetilde{\sigma}}{M} \right )^2 $$ $z$ = z multiplier for given confidence level $\widetilde{\sigma}$ = estimated population standard deviation $M$ = margin of error ::: The sample standard deviation may be estimated on the basis of prior research studies. ::: {#exm-estimating-iq} #### Estimating IQ A team of researchers wants to estimate the mean IQ of students enrolled at one prestigious university. Previous research studies have examined samples of students from other similar universities and usually find results around $\bar{x}=120$ and $s=10$. In order to construct a 90% confidence interval with a margin of error of $\pm2$ IQ points, what sample size should be obtained? As shown in the probability distribution plot below, the z value associated with a 90% confidence interval is 1.645. ![](assets/Multiplierfor90CI.png){#fig-Multiplierfor90CI fig-alt="Standard normal distribution showing the z multipliers for a 90% confidence interval" .mx-auto .d-block width="60%" .lightbox} The estimated standard deviation is given to be 10 and the desired margin of error is given to be 2. $$ n=\dfrac{z^{2}\widetilde{\sigma}^{2}}{M^{2}}=\dfrac{1.645^{2}(10^{2})}{2^{2}}=67.615 $$ We round up to 68. The research team should attempt to obtain a sample of at least 68 individuals. ::: ::: {#exm-estimating-age} ##### Video: Age ::: {.w-75 .mx-auto #vid-8-5} {{< video https://youtu.be/LU4i5delpB4?si=tgX7RFvGQE9H4CnM >}} STAT 200 Estimating Necessary Sample Size for CI of Mu (Age Example) ::: ::: ::: {#exm-estimating-cookie} ##### Video: Cookie Weights ::: {.w-75 .mx-auto #vid-8-6} {{< video https://youtu.be/jAW_ZYGQcqI?si=Pcd4EGfb5s6-swvJ >}} N to Estimate Mean Cookie Weight ::: ::: ### Hypothesis Testing In this section, we will be comparing one sample mean to one known or hypothesized population value. In [Lesson 5](Lesson05.qmd) you learned how to conduct randomization tests. Here, you will learn how to conduct a one sample mean t-test and a one sample mean z-test. The t-distribution is used to estimate the sampling distribution when the sample size is large (at least 30) or when the population is known to be normally distributed (but $\sigma$ is unknown). The z-distribution is used on rare occasions when the population is normal and the population standard deviation is known. Note that for this course the one sample mean z-test is optional; it used only in specific cases where the population is known to be normally distributed and when the population standard deviation ($\sigma$) is known. The most commonly used one sample mean test is the "one sample mean t-test" which is also known as a "single sample mean t-test." ![Flow Chart: Approximating the sample distribution](/assets/one-sample-mean-flowchart.png){#fig-one-sample-mean-flowchart.png2 fig-alt="Flowchart of z- and t- Distributions. Description below." fig-align="center" width="70%" .lightbox} <details> <summary>Text alternative of @fig-one-sample-mean-flowchart.png2</summary> - Is the population known to be normally distributed? - Yes: - Is the population standard deviation known? - Yes: z-distribution. - No: t-distribution. - No: - Is the sample size at least 30? - Yes: t-distribution. - No: Bootstrap/Randomization. </details> #### One Sample Mean t-Test, Formulas ##### Five-Step Hypothesis Testing Procedure {.unnumbered .unlisted} ::: {.steps} 1. **Check assumptions and write hypotheses.** Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large ($\ge\ 30$) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are: | Research Question | Is the mean different from $\mu_{0}$? | Is the mean greater than $\mu_{0}$? | Is the mean less than $\mu_{0}$? | | --- |--- |--- |--- | | Null Hypothesis, $H_{0}$ | $\mu=\mu_{0}$ | $\mu=\mu_{0}$ | $\mu=\mu_{0}$ | | Alternative Hypothesis, $H_{a}$ | $\mu\neq \mu_{0}$ | $\mu> \mu_{0}$ | $\mu<\mu_{0}$ | | Type of Hypothesis Test | Two-tailed, non-directional | Right-tailed, directional | Left-tailed, directional | : {.w-auto .table-sm .mx-auto .row-header} where $\mu_{0}$ is the hypothesized population mean. 2. **Calculate the test statistic.** For the test of one group mean we will be using a t-test statistic: ::: {.callout-note appearance="minimal"} [**Test Statistic: One Group Mean**]{.fs-4} $$ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}} $$ {#eq-test-statistic-one-group-mean} $\bar{x}$ = sample mean $\mu_{0}$ = hypothesized population mean $s$ = sample standard deviation $n$ = sample size ::: Note that structure of this formula is similar to the general formula for a test statistic: $$ \dfrac{\text{sample statistic}-\text{null value}}{\text{standard error}} $$ 3. **Determine the p-value.** When testing hypotheses about a mean or mean difference, a t-distribution is used to find the $p$-value. These t-distributions are indexed by a quantity called degrees of freedom, calculated as $df = n – 1$ for the situation involving a test of one mean or test of mean difference. The $p$-value can be found using Minitab. 4. **Make a decision.** If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. 5. **State a "real world" conclusion.** Based on your decision in Step 4, write a conclusion in terms of the original research question. ::: Next, you will go through a few examples, and then you will have the opportunity to do two on your own. :::{#exm-book-costs} #### Book Costs **Research question:** Does the average Penn State student spend more than \$300 each semester on textbooks? In a sample of 226 Penn State students, the mean cost of a student’s textbooks was \$344 with a standard deviation of \$106. ::: {.w-75 .mx-auto #vid-8-7} {{< video https://youtu.be/C3G0VcwV7PE?si=uB91McrtSSh3uxDQ >}} STAT 200 Single Sample Mean t Hypothesis Test Example: Textbook Spending ::: ::: :::{#exm-pulse-rate} #### Pulse Rate A research study measured the pulse rates of 57 college men and found a mean pulse rate of 70.4211 beats per minute with a standard deviation of 9.9480 beats per minute. Researchers want to know if the mean pulse rate for all college men is different from the current standard of 72 beats per minute. ::: {.steps} 1. **Check assumptions and write hypotheses.** Pulse rates are quantitative. The sampling distribution will be approximately normally distributed because $n \ge 30$. This is a two-tailed test because we want to know if the mean pulse rate is **different** from 72. $H_{0}:\mu=72$ $H_{a}: \mu\neq 72$ 2. **Calculate the test statistic.**(@eq-test-statistic-one-group-mean) $$ \bar{x}=70.4211, \mu_0=72, s=9.9480, n=57 $$ $$ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{70.4211-72}{\dfrac{9.9480}{\sqrt{57}}}=-1.198 $$ Our t-test statistic is -1.198. 3. **Determine the p-value.** $$ df=n-1=57-1=56 $$ ![](assets/t56twotail1198.png){#fig-t56twotail1198 fig-alt="Distribution Plot of Density vs X - T, DF=56" .mx-auto .d-block width="60%" .lightbox} $$ p=0.117981+0.117981=0.235962 $$ Given that the null hypothesis is true and $\mu=72$, the probability of taking a random sample of $n=57$ and finding a sample mean this or more extremely different is 0.235962. This is our p-value. 4. **Make a decision.** $p>.05$, therefore we fail to reject the null hypothesis. 5. **State a "real world" conclusion.** There is not sufficient evidence to state that the mean pulse of college men is different from 72. ::: ::: :::{#exm-coffee} #### Coffee In the population of Americans who drink coffee, the average daily consumption is 3 cups per day. A university wants to know if their students tend to drink more coffee than the national average. They ask a random sample of 50 students how many cups of coffee they drink each day and found $\bar{x}=3.8$ and $s=1.5$. Do they have convincing evidence that their students drink more than the national average? ::: {.steps} 1. **Check assumption and write hypotheses.** Amount of coffee consumed is a quantitative variable. We are given that random sampling methods were employed. Because $n \ge 30$, we can approximate the sampling distribution using a t distribution. This is a right-tailed test because we want to know if the mean in the sample is **greater than** the national average. $H_{0}:\mu= 3$ $H_{a}:\mu>3$ 2. **Calculate the test statistic.** (@eq-test-statistic-one-group-mean) $$ \bar{x}=3.8, \mu_0=3, s=1.5, n=50 $$ $$ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{3.8-3}{\dfrac{1.5}{\sqrt{50}}}=3.771 $$ Our t-test statistic is 3.771 3. **Determine the p-value.** $df=n-1=50-1=49$ ![](assets/t3.771.png){#fig-t3.771 fig-alt="Distribution Plot of Density vs X - T, DF=49" .mx-auto .d-block width="60%" .lightbox} Using Minitab, we can find that $P(t > 3.771) =0.0002191$. p-value = 0.0002191 If $\mu=3$, then the probability of taking a random sample of $n=50$ and finding $\bar{x} \geq 3$ is 0.0002191 4. **Make a decision.** $p\leq.05$, therefore we reject the null hypothesis. 5. **State a "real world" conclusion.** There is convincing evidence to state the mean number of coffees consumed in the population of all students at this university is greater than 3. ::: ::: :::{#exm-transportation-costs} #### Transportation Costs According to [CNN](http://money.cnn.com/interactive/news/economy/us-spending), in 2011, the average American spent \$16,803 on housing. A suburban community wants to know if their residents spent less than this national average. In a survey of 30 randomly selected residents, they found that they spent an annual average of \$15,800 with a standard deviation of \$2,600. ::: {.steps} 1. **Check assumptions and write hypotheses.** Housing costs are quantitative. Because $n \ge 30$, the sampling distribution can be approximated using the t-distribution. This is a left-tailed test because we want to know if residents of this community spent **less than** the national average. $H_{0}:\mu=16803$ $H_{a}:\mu<16803$ 2. **Calculate the test statistic.** (@eq-test-statistic-one-group-mean) $$ \bar{x}=15800, \mu_0=16803, s=2600, n=30 $$ $$ t=\dfrac{\bar{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{15800-16803}{\dfrac{2600}{\sqrt{30}}}=-2.113 $$ Our t-test statistics is -2.113. 3. **Determine the p-value.** $$ df=n-1=30-1=29 $$ This is a left-tailed test so we want to know the probability of $t < -2.113$. ![](assets/tlessthanneg2.113.png){#fig-tlessthanneg2.113 fig-alt="Distribution Plot of Density vs X - T, DF=29" .mx-auto .d-block width="60%" .lightbox} Using Minitab we can find that $p=0.0216634$. 4. **Make a decision.** $p\leq .05$, therefore we reject the null hypothesis. 5. **State a "real world" conclusion.** There is convincing evidence to state that on average residents of this community spent less than the national average on housing in 2011. ::: ::: #### ![ ](assets/minitab.png){.icon-image} Minitab: One Sample Mean t-Tests A hypothesis test for one group mean can be conducted in Minitab using raw data or summarized data. - If you have a data file with every individual's observation, then you have **raw data**. - If you do not have each individual's observation, but rather have the sample mean, sample standard deviation, and sample size, then you have **summarized data**. {{< include _minitab/conduct-a-one-sample-t-hypothesis-test-with-raw-data.qmd >}} {{< include _minitab/conduct-a-one-sample-t-hypothesis-test-with-summarized-data.qmd >}} The next two examples will show you how to use Minitab to conduct a one-sample mean t-test using either raw data or summarized data. There is also one example of using Minitab to conduct a one-sample mean z-test which is only performed if the population is known to be normally distributed and the population standard deviation ($\sigma$) is available. ::: {#exm-minitab-raw-data-onesamplet .example-minitab} #### Minitab: One Sample Mean t-Test Using Raw Data **Research question**: Is the mean GPA in the population different from 3.0? The GPAs of 226 students are available. ::: {.steps} 1. **Check assumptions and write hypotheses.** We do not know whether the population is normally distributed. However, the sample size is large ($n=226 \ge 30$), so the sampling distribution of the sample mean is approximately normal and we can perform a one-sample mean t-test. $H_0\colon \mu = 3.0$ $H_a\colon \mu \ne 3.0$ 2. **Calculate the test statistic.** Use Minitab to conduct the one-sample mean t-test with raw data: 1. Open the Minitab file: ([class_survey.mpx](data_files/class_survey.mpx){download="" target="_blank"}). 2. Select **Stat > Basic Statistics > 1-Sample t**. 3. Select **One or more samples, each in a column** from the dropdown. 4. Insert the variable *GPA* into the **Sample** box. 5. Check **Perform a hypothesis test**. 6. Enter *3* for the **Hypothesized mean**. 7. Select **Options**. 8. Use the default alternative hypothesis, **Mean ≠ hypothesized mean**. 9. Use the default confidence level of 95. 10. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} | N | Mean | StDev | SE Mean | 95% CI for $\mu$ | | --- |--- |--- |--- |--- | | 226 | 3.2311 | 0.5104 | 0.0340 | (3.1642, 3.2980) | : {.w-auto .table-sm .row-header} *$\mu$: population mean of GPA* ##### Test **Null hypothesis:** $H_0\colon \mu = 3$ **Alternative hypothesis:** $H_1\colon \mu \ne 3$ | T-Value | P-Value | | --- |--- | | 6.81 | 0.000 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, $t(225)=6.81$. 3. **Determine the p-value.** Minitab reports the p-value as 0.000. We write this as $p<0.001$. 4. **Make a decision.** $p \le \alpha$, so we reject the null hypothesis. 5. **State a "real world" conclusion.** There is convincing evidence that the mean GPA in the population is different from 3.0. ::: ::: ::: {#exm-minitab-raw-data-onesampletsumm .example-minitab} #### Minitab: One Sample Mean t-Test Using Summarized Data Here we are testing $H_{a}\colon\mu\neq72$ and are given $n=35$, $\bar{x}=76.8$, and $s=11.62$. ::: {.steps} 1. **Check assumptions and write hypotheses.** The shape of the population distribution is unknown. However, because $n \ge 30$, we can perform a one-sample mean *t*-test. $H_0\colon \mu = 72$ $H_a\colon \mu \ne 72$ 2. **Calculate the test statistic.** Use Minitab to conduct the one-sample mean *t*-test with summarized data: 1. Select **Stat > Basic Statistics > 1-Sample t**. 2. Select **Summarized data** from the dropdown. 3. Enter **35** for the sample size, **76.8** for the sample mean, and **11.62** for the standard deviation. 4. Select **Perform a hypothesis test**. 5. Enter **72** for the hypothesized mean. 6. Select **Options**. 7. Use the default alternative hypothesis of **Mean ≠ hypothesized mean**. 8. Use the default confidence level of 95. 9. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Descriptive Statistics {.unnumbered .unlisted} | N | Mean | StDev | SE Mean | 95% CI for $\mu$ | | --- |--- |--- |--- |--- | | 35 | 76.80 | 11.62 | 1.96 | (72.81, 80.79) | : {.w-auto .table-sm .row-header} $\mu$: population mean of Sample ##### Test {.unnumbered .unlisted} **Null hypothesis:** $H_0\colon \mu = 72$ **Alternative hypothesis:** $H_1\colon \mu \ne 72$ | T-Value | P-Value | | --- |--- | | 2.44 | 0.0199 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, the test statistic is $t(34)=2.44$. 3. **Determine the p-value.** From the output, $p=0.0199$. 4. **Make a decision.** Because $p \le \alpha$, we reject the null hypothesis. 5. **State a "real world" conclusion.** There is convincing evidence that the population mean is different from 72. ::: ::: #### One Sample Mean z-Test (Optional) A one sample mean z-test is used when the population is known to be normally distributed and when the population standard deviation ($\sigma$) is known. This most frequently occurs in the social sciences when standardized measures are used such as IQ, SAT, ACT, or GRE scores, for which the population parameters are known. The formula for computing a z-test statistic for one sample mean is identical to that of computing a t-test statistic for one sample mean, except now the population standard deviation is known and can be used in computing the standard error. ::: {.callout-note appearance="minimal"} [**z-Test Statistic: One Group Mean**]{.fs-4} $$ z=\dfrac{\bar{x}-\mu_0}{\dfrac{\sigma}{\sqrt{n}}} $$ $\bar{x}$ = sample mean $\mu_{0}$ = hypothesized population mean $s$ = sample standard deviation $n$ = sample size ::: The other primary difference between the one sample mean t-test and the one sample mean z-test is the latter uses the standard normal distribution (i.e., z-distribution) in determining the p-value. Below are the directions for conducting a one sample mean z-test in Minitab. ::: {#exm-minitab-raw-data-onesamplez .example-minitab} #### Minitab: Performing a One Sample Mean z-Test **Research question:** Are the IQ scores of students at one college-prep school above the national average? Scores on one American IQ test are normed to have a mean of 100 and standard deviation of 15. In a simple random sample of 25 students at this school the mean was 110. ::: {.steps} 1. **Check assumptions and write hypotheses.** The population is known to be normally distributed, and the population standard deviation is known to be 15. With these two conditions met, we can conduct a one-sample mean z-test. $H_0\colon \mu = 100$ $H_a\colon \mu > 100$ 2. **Calculate the test statistic.** Use Minitab to conduct the one-sample mean z-test with summarized data: 1. Select **Stat > Basic Statistics > 1-Sample Z**. 2. Select **Summarized data** from the dropdown. 3. Enter 25 for the sample size, 110 for the sample mean, and 15 for the known standard deviation. 4. Check **Perform a hypothesis test**. 5. Enter 100 for the **Hypothesized mean**. 6. Select **Options**. 7. Set the alternative hypothesis to **Mean > hypothesized mean**. 8. Use the default confidence level of 95. 9. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} ##### Descriptive Statistics | N | Mean | SE Mean | 95% Lower Bound for $\mu$ | | --- |--- |--- |--- | | 25 | 110.00 | 3.00 | 105.07 | : {.w-auto .table-sm .row-header} *$\mu$: population mean of Sample* *Known standard deviation = 15* ##### Test **Null hypothesis:** $H_0\colon \mu = 100$ **Alternative hypothesis:** $H_1\colon \mu > 100$ | Z-Value | P-Value | | --- |--- | | 3.33 | 0.000 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, $z=3.33$. 3. **Determine the p-value.** Minitab reports the p-value as 0.000. We write this as $p<0.001$. 4. **Make a decision.** $p \le \alpha$, so we reject the null hypothesis. 5. **State a "real world" conclusion.** There is convincing evidence that the mean IQ score of all students at this school is greater than 100. ::: ::: ## Paired Means In [Lesson 1](Lesson01.qmd) we learned about independent samples and paired samples. When we have two independent samples, the observations in the two groups are unrelated to one another and are not matched in any meaningful way. We'll learn how to compare the means of two independent groups in [Lesson 9](Lesson09.qmd). With paired samples, the observations in the two groups are matched in a meaningful way. These are also known as dependent samples. Most often this occurs when data are collected twice from the same participants, called repeated measures. For example, think of studying the effectiveness of a diet plan. You would weigh each participant prior to starting the diet and again following some time on the diet. Depending on how much weight they lost you would determine if the diet was effective. Paired data does not always need to involve two measurements on the same subject; it can also involve taking one measurement on each of two related subjects. For example, we may study husband-wife pairs, mother-son pairs, or pairs of twins. In constructing a dependent samples confidence interval or conducting a dependent samples hypothesis test, the difference score is computed for each individual or pair. From there, the procedures are the same that you used for constructing confidence intervals and hypothesis tests for single sample means. As with one sample mean, if the sample size is at least 30, the sampling distribution for the difference in paired means can be approximated using a t-distribution. In terms of symbols, the population parameter of interest is the mean difference in the population "$\mu_d$." This is estimated using the mean difference in the sample "$\bar x_d.$" ### Confidence Intervals Recall the general form of a confidence interval... $$ \text{sample statistic}\pm\underbrace{(\text{multiplier})\ (\text{standard error})}_{\textbf{margin of error}}. $$ The formula for constructing a confidence interval for the difference in paired means is almost identical to the formula for constructing a confidence interval for one mean. Note that the only change is the subscript d which stands for difference. ::: {.callout-note appearance="minimal"} [**Confidence Interval for the Difference Between Two Paired Means**]{.fs-4} $$ \underbrace{\bar{x}_d}_{\text{sample statistic}} \pm \overbrace{t^*}^{\text{multiplier}} \underbrace{\left(\dfrac{s_d}{\sqrt{n}}\right)}_{\text{standard error}} $$ $t^*$ is the multiplier with $df = n-1$. ::: ::: {#exm-change-in-knowledge} #### Change in Knowledge An educational research study is designed so that participants complete a measure of demonstrated knowledge twice. The researcher wants to estimate the change in scores from the first to second administrations (i.e., pre- and post-test). Data are paired by participant. The researcher subtracted pre-test scores from the post test scores and found a mean increase of 6.560 with a standard deviation of 3.867 for $n=100$. She wants to construct a 95% confidence interval for the mean difference. First, we'll find the appropriate multiplier. $$ df=n-1=100-1=99 $$ ![](assets/tMultdf99level95.png){#fig-tMultdf99level95 fig-alt="t distribution showing the multipliers for a 95% confidence interval given 99 degrees of freedom." .mx-auto .d-block width="60%" .lightbox} For a 95% confidence interval: $t_{df=99}=1.984$ $$ 6.560 \pm 1.984 \left(\frac{3.867}{\sqrt{100}}\right)=6.560 \pm 0.767=[5.793, 7.327] $$ We are 95% confident that the mean difference between post- and pre-test scores in the population is between 5.793 and 7.327. Data from Zimmerman, W. A. (2015). Impact of Instructional Materials Eliciting Low and High Cognitive Load on Self-Efficacy and Demonstrated Knowledge (Unpublished doctoral dissertation). The Pennsylvania State University, University Park, PA. ::: ::: {#exm-viddiffexamscores} #### Video: Difference in Exam Scores ::: {.w-75 .mx-auto #vid-8-8} {{< video https://youtu.be/UMgHYwpQosM?si=qeD_i-HiFlMWv8Mk >}} STAT 200 95% CI for Mean Difference in Exam Scores ::: ::: #### ![ ](assets/minitab.png){.icon-image} Minitab: Paired Means Confidence Interval {{< include _minitab/construct-a-paired-t-confidence-interval.qmd >}} ### Hypothesis Testing Below are the procedures for conducting a hypothesis test for two paired means. This is often referred to as a "paired means t-test," "dependent means t-test," or "matched pairs t-test." ::: {.steps} 1. **Check any necessary assumptions and write null and alternative hypotheses.** Data must be paired. The difference between the two groups must be normally distributed in the population or the sample size must be at least 30. The possible combinations of null and alternative hypotheses are: | Research Question | Is the mean difference different from 0? | Is the mean difference greater than 0? | Is the mean difference less than 0? | | --- |--- |--- |--- | | Null Hypothesis, $H_{0}$ | $\mu_d = 0$ | $\mu_d = 0$ | $\mu_d = 0$ | | Alternative Hypothesis, $H_{a}$ | $\mu_d \neq 0$ | $\mu_d > 0$ | $\mu_d < 0$ | | Type of Hypothesis Test | Two-tailed, non-directional | Right-tailed, directional | Left-tailed, directional | : {.w-auto .table-sm .mx-auto .row-header} Where $\mu_d$ is the hypothesized difference in the population. 2. **Calculate an appropriate test statistic.** The calculation of the test statistic for dependent samples is similar to the calculation you performed earlier in this lesson for a single sample mean. In this formula, $\bar{x}_d$ is used in place of $\bar{x}$ and $s_d$ is used in place of $s$: ::: {.callout-note appearance="minimal"} [**Test Statistic for Dependent Means**]{.fs-4} $$ t=\dfrac{\bar{x}_d-\mu_0}{\dfrac{s_d}{\sqrt{n}}} $$ - $\bar{x}_d$ = observed sample mean difference - $\mu_0$ = mean difference specified in the null hypothesis - $s_d$ = standard deviation of the differences - $n$ = sample size (i.e., number of unique individuals) ::: ::: {.callout-note appearance="minimal"} [**Observed Sample Mean Difference**]{.fs-4} $$ \bar{x}_d=\dfrac{\Sigma{x}_d}{n} $$ - $x_d$ = observed difference ::: ::: {.callout-note appearance="minimal"} [**Standard Deviation of the Differences**]{.fs-4} $$ s_d=\sqrt{\dfrac{\sum (x_d-\bar{x}_d)^{2}}{n-1}} $$ ::: 3. **Determine the p-value associated with the test statistic.** When testing hypotheses about a mean difference, a t-distribution is used to find the $p$ value. The degrees of freedom are equal to $n-1$ where $n$ is the number of pairs. 4. **Decide between the null and alternative hypotheses.** If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. 5. **State a "real world" conclusion.** Based on your decision in Step 4, write a conclusion in terms of the original research question. ::: ::: {#exm-quiz-scores} #### Quiz Scores Below is an example of conducting a paired means t-test by hand using raw data. Next, you will learn how this can be conducted most efficiently in Minitab. **Research question:** Are scores on two quizzes different? Data were collected from 9 students and a paired means t-test was performed using hand calculations: | Student ID | Quiz 1 | Quiz 2 | | --- |--- |--- | | 001 | 98 | 94 | | 002 | 100 | 98 | | 003 | 95 | 98 | | 004 | 90 | 88 | | 005 | 90 | 89 | | 006 | 92 | 91 | | 007 | 80 | 84 | | 008 | 78 | 80 | | 009 | 88 | 88 | : {.w-auto .table-sm .mx-auto .row-header} ::: {.steps} 1. **Check assumptions and write hypotheses.** There are two assumptions: (1) data are paired and (2) distribution of differences is normally distribution in the population or the sample size is at least 30. The data are paired because for each student we have a quiz 1 and a quiz 2 score. We do not know if the differences are normally distributed in the population and the sample size is small, but in the video above we created a histogram of the differences and found that the sample was approximately normally distributed, so this assumption has been met, and we can perform a paired means t-test. Given $\mu_d = \mu_1 - \mu_2$, our hypotheses are: $H_0: \mu_d = 0$ $H_a: \mu_d \ne 0$ 2. **Calculate test statistic.** | Student ID | Quiz 1 | Quiz 2 | Difference ($X_d$) | $X_d - \bar{X}_d$ | $(X_d - \bar{X}_d)^2$ | | --- |--- |--- |--- |--- |--- | | 001 | 98 | 94 | 4 | 3.889 | 15.123 | | 002 | 100 | 98 | 2 | 1.889 | 3.568 | | 003 | 95 | 98 | -3 | -3.111 | 9.679 | | 004 | 90 | 88 | 2 | 1.889 | 3.568 | | 005 | 90 | 89 | 1 | 0.889 | 0.790 | | 006 | 92 | 91 | 1 | 0.889 | 0.790 | | 007 | 80 | 84 | -4 | -4.111 | 16.901 | | 008 | 78 | 80 | -2 | -2.111 | 4.457 | | 009 | 88 | 88 | 0 | -0.111 | 0.012 | : {.w-auto .table-sm .mx-auto .row-header} Mean of the differences: $\bar{X}_d=\dfrac{\Sigma{X}_d}{n}=\dfrac{1}{9}$ For a review of computing standard deviation, see [Lesson 2](Lesson02.qmd). Sum of squares: $\Sigma (X_d - \bar{X}_d)^2 = 54.889$ Standard deviation of the differences: $s_d=\sqrt{\dfrac{\sum (X_d-\bar{X}_d)^{2}}{n-1}} = \sqrt{\dfrac{54.889}{9-1}}=2.619$ Test statistic: $t=\dfrac{\bar{X}_d- \mu_0}{\dfrac{s_d}{\sqrt{n}}}=\dfrac{\dfrac{1}{9}}{\dfrac{2.619}{\sqrt{9}}}=0.127$ $df=n-1=9-1=8$ 3. **Determine p-value.** We can construct a t-distribution with 8 degrees of freedom and determine what proportion of the curve falls beyond a $t$ score of 0.127. This is a two-tailed test, so we need to take into account both the left and right sides of the curve. ![](assets/tdist0.127twotailed.png){#fig-tdist0.127twotailed fig-alt="Distribution Plot of Density vs X - T, df=8" .mx-auto .d-block width="60%" .lightbox} $$ p=0.4510+0.4510=0.9020 $$ 4. **Make a decision.** We will compare our $p$-value from step 3 to a standard alpha level of 0.05. Because $p>\alpha$, we fail to reject the null hypothesis. 5. **State conclusion.** There is not sufficient evidence to state that scores on the two quizzes are different. ::: ::: {.callout-caution appearance="minimal"} **Note!** \ The following video uses Minitab Express not Minitab to find the p-value ::: ::: {.w-75 .mx-auto #vid-8-9} {{< video https://youtu.be/Dp8-cC0vld4?si=rNr_CFFjM8QZ1tuH >}} Paired Means t Test Quiz Scores - Hand Calculations Example ::: ::: #### ![ ](assets/minitab.png){.icon-image} Minitab: Paired Means Test The steps for constructing a confidence interval or conducting a paired means *t*-test in Minitab are identical. The output that the procedure provides includes both the confidence interval and the $p$-value for determining statistical significance. {{< include _minitab/conduct-a-paired-t-hypothesis-test.qmd >}} ::: {#exm-minitab-pairedmeanssat .example-minitab} #### Minitab: Conducting a Paired Means Test Let's compare students' SAT-Math scores to their SAT-Verbal scores. **Research question:** Do students score differently on the SAT-Math and SAT-Verbal tests? ::: {.steps} 1. **Check assumptions and write hypotheses.** Because the sample size is large ($n \ge 30$), the t-distribution may be used to approximate the sampling distribution. $H_{0}:\mu_d=0$ $H_{a}:\mu_d \ne 0$ 2. **Calculate the test statistic.** Use Minitab to conduct the paired means test: 1. Open the Minitab file: ([class_survey.mpx](data_files/class_survey.mpx){download="" target="_blank"}). 2. Select **Stat > Basic Statistics > Paired t**. 3. Select **Each sample is in a column** since the data are in the worksheet. 4. Insert the variable *SATM* into the **Sample 1** box. 5. Insert the variable *SATV* into the **Sample 2** box. 6. Select **OK**. This should result in the following output: ::: {.minitab_output} #### Paired t: SATM, SATV {.unnumbered .unlisted} ##### Descriptive Statistics | Sample | N | Mean | StDev | SE Mean | | --- |--- |--- |--- |--- | | SATM | 215 | 599.81 | 84.70 | 5.78 | | SATV | 215 | 580.33 | 82.44 | 5.62 | : {.w-auto .table-sm .row-header} ##### Estimation for Paired Difference | Mean | StDev | SE Mean | 95% CI for $\mu_d$ | | --- |--- |--- |--- | | 19.49 | 89.81 | 6.12 | (7.42, 31.56) | : {.w-auto .table-sm .row-header} $\mu_{\text{difference}}$: population mean of (SATM - SATV) ##### Test **Null hypothesis:** $H_0\colon \mu_{\text{difference}} = 0$ **Alternative hypothesis:** $H_1\colon \mu_{\text{difference}} \ne 0$ | T-Value | P-Value | | --- |--- | | 3.18 | 0.002 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, the t-test statistic is 3.18. 3. **Determine the p-value associated with the test statistic.** From the output, the p-value is 0.002. 4. **Make a decision.** $p\leq .05$; therefore, our decision is to reject the null hypothesis. 5. **State a "real world" conclusion.** There is evidence that in the population, on average, students' SAT-Math and their SAT-Verbal scores are different. ::: ::: ### <i class="bi bi-question-octagon-fill text-warning"></i> Try It! Marriage Age (Summarized Data){.unnumbered .unlisted} ::: {.tryit} In a sample of 105 married heterosexual couples, the average age difference (husband's age - wife's age) was 2.829 years with a standard deviation of 4.995 years. These summary statistics were taken from a data set from the Lock<sup>5</sup> textbook. Is there convincing evidence that, on average, in the population, husbands tend to be older than their wives? ```{=html} <button id="header-4" class="btn btn-outline-success collapsed" type="submit" data-bs-toggle="collapse" data-bs-target="#panel-4" aria-expanded="false" aria-controls="panel-4"> Answer </button> ``` ::: {.collapse #panel-4 aria-labelledby="header-4"} ::: {.card .card-body} ::: {.steps} 1. **Check assumptions and write hypotheses.** Because the sample size is large ($n \ge 30$), the *t* distribution may be used to approximate the sampling distribution. Let $\mu_d$ be the population mean difference in age (husband's age minus wife's age). $H_{0}:\mu_d=0$ $H_{a}:\mu_d \gt 0$ 2. **Calculate the test statistic.** Use Minitab to conduct the paired means test using the summarized differences: 1. Select **Stat > Basic Statistics > Paired t**. 2. Select **Summarized data (differences)**. 3. Enter **105** for the sample size, **2.829** for the sample mean, and **4.995** for the standard deviation. 4. Select **Options**. 5. Enter **0** for the hypothesized difference. 6. For the alternative hypothesis, select **Difference > hypothesized difference**. 7. Select **OK**, then **OK**. This should result in the following output: ::: {.minitab_output} #### Estimation for Paired Difference | Mean | StDev | SE Mean | 95% CI for $\mu_d$ | | --- |--- |--- |--- | | 2.829 | 4.995 | 0.487 | (1.862, 3.796) | : {.w-auto .table-sm .row-header} $\mu_{\text{difference}}$: population mean of (husband's age - wife's age) ##### Test **Null hypothesis:** $H_0\colon \mu_{\text{difference}} = 0$ **Alternative hypothesis:** $H_1\colon \mu_{\text{difference}} > 0$ | T-Value | P-Value | | --- |--- | | 5.80 | 0.000 | : {.w-auto .table-sm .row-header} ::: From the Minitab output, the t-test statistic is 5.80. 3. **Determine the p-value associated with the test statistic.** From the output, the p-value is reported as 0.000. We write this as $p < 0.001$. 4. **Make a decision.** $p\leq .05$; therefore, our decision is to reject the null hypothesis. 5. **State a "real world" conclusion.** There is convincing evidence that in the population, on average, the husband's age in heterosexual couples is greater than the wife's age. ::: ::: ::: ::: ## Summary In this lesson, you learned how to construct confidence intervals and conduct hypothesis tests for one proportion, one mean, and paired means using the z- and t-distributions. These procedures are sometimes referred to as "traditional" methods, in contrast to the bootstrapping and randomization procedures that we learned earlier in the course which are "simulation-based inference" methods. The table below summarizes the procedures covered in this lesson. Note the similarities between all the confidence interval formulas and all of the test statistic formulas. These all follow the same general format. | Procedure | Assumptions | Standard Error | Confidence Interval | Test Statistic | | --- |--- |--- |--- |--- | | One Proportion $z$ | Confidence interval: At least 10 successes and 10 failures in the sampleHypothesis test: $n p_0 \ge 10$ and $n (1-p_0) \ge 10$ | $\sqrt{\dfrac{p(1-p)}{n}}$ | $\widehat{p} \pm z^{*} \left ( \sqrt{\dfrac{\hat{p} (1-\hat{p})}{n}} \right)$ | $z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}$ | | One Mean $t$ | Sample size is at least 30 OR population is normally distributedPopulation standard deviation is unknown* | $\dfrac{\sigma}{\sqrt n}$ | $\bar{x} \pm t^{*} \dfrac{s}{\sqrt{n}}$ | $t=\dfrac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}$ | | Paired Means $t$ | Sample size is at least 30 OR differences are normally distributed in the population | $\dfrac{\sigma_d}{\sqrt n}$ | $\bar{x}_d \pm t^* \left(\dfrac{s_d}{\sqrt{n}}\right)$ | $t=\dfrac{\bar{x}_d-\mu_0}{\frac{s_d}{\sqrt{n}}}$ | : {.w-auto .table-sm .mx-auto .row-header} \* If the population standard deviation is known, then the sampling distribution may be approximated by a z-distribution. This lesson may be overwhelming if you are trying to memorize all the formulas that were introduced. Luckily, doing so is unnecessary. You should focus on seeing the similarities between all the procedures. For one sample mean, one sample proportion, paired means, two independent means, and two independent proportions, the following general formulas can be applied ::: {.callout-note appearance="minimal"} [**General Form of a Confidence Interval**]{.fs-4} $$ \text{sample statistic}\pm(\text{multiplier})\ (\text{standard error}) $$ ::: ::: {.callout-note appearance="minimal"} [**General Form of a Test Statistic**]{.fs-4} $$ \text{test statistic}=\dfrac{\text{sample statistic}-\text{null parameter}}{\text{standard error}} $$ ::: Also, when given raw data, use Minitab. It is much more efficient than performing calculations by hand.

Research Question	Is the proportion different from \(p_0\)?	Is the proportion greater than \(p_0\)?	Is the proportion less than \(p_0\)?
Null Hypothesis, \(H_{0}\)	\(p=p_0\)	\(p= p_0\)	\(p= p_0\)
Alternative Hypothesis, \(H_{a}\)	\(p\neq p_0\)	\(p> p_0\)	\(p< p_0\)
Type of Hypothesis Test	Two-tailed, non-directional	Right-tailed, directional	Left-tailed, directional

Research Question	Is the mean different from \(\mu_{0}\)?	Is the mean greater than \(\mu_{0}\)?	Is the mean less than \(\mu_{0}\)?
Null Hypothesis, \(H_{0}\)	\(\mu=\mu_{0}\)	\(\mu=\mu_{0}\)	\(\mu=\mu_{0}\)
Alternative Hypothesis, \(H_{a}\)	\(\mu\neq \mu_{0}\)	\(\mu> \mu_{0}\)	\(\mu<\mu_{0}\)
Type of Hypothesis Test	Two-tailed, non-directional	Right-tailed, directional	Left-tailed, directional

Procedure	Assumptions	Standard Error	Confidence Interval	Test Statistic
One Proportion \(z\)	Confidence interval: At least 10 successes and 10 failures in the sampleHypothesis test: \(n p_0 \ge 10\) and \(n (1-p_0) \ge 10\)	\(\sqrt{\dfrac{p(1-p)}{n}}\)	\(\widehat{p} \pm z^{*} \left ( \sqrt{\dfrac{\hat{p} (1-\hat{p})}{n}} \right)\)	\(z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}\)
One Mean \(t\)	Sample size is at least 30 OR population is normally distributedPopulation standard deviation is unknown*	\(\dfrac{\sigma}{\sqrt n}\)	\(\bar{x} \pm t^{*} \dfrac{s}{\sqrt{n}}\)	\(t=\dfrac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}\)
Paired Means \(t\)	Sample size is at least 30 OR differences are normally distributed in the population	\(\dfrac{\sigma_d}{\sqrt n}\)	\(\bar{x}_d \pm t^* \left(\dfrac{s_d}{\sqrt{n}}\right)\)	\(t=\dfrac{\bar{x}_d-\mu_0}{\frac{s_d}{\sqrt{n}}}\)

Objectives

8.1 One Sample Proportion

8.1.1 Confidence Intervals

Normal Approximation Formulas

Finding the \(z^{*}\) Multiplier

Minitab: Confidence Interval for a Proportion

Method

Descriptive Statistics

Try It!

Method

Method

Descriptive Statistics

8.1.2 Computing Necessary Sample Size

8.1.3 Hypothesis Testing

Normal Approximation Method Formulas

Minitab: Hypothesis Tests for One Proportion

Minitab: 1 Proportion z-Test, Raw Data

Method

Descriptive Statistics

Test

Minitab: 1 Proportion z-Test, Summarized Data

Method

Descriptive Statistics

Test

Method

Descriptive Statistics

Test

8.2 One Sample Mean

8.2.1 t-Distribution

Minitab: Using t-Distribution Probability Plots

8.2.2 Confidence Intervals

Formulas

Descriptive Statistics

Try It!

Descriptive Statistics

Descriptive Statistics

8.2.3 Computing Necessary Sample Size

8.2.4 Hypothesis Testing

One Sample Mean t-Test, Formulas

Five-Step Hypothesis Testing Procedure

Minitab: One Sample Mean t-Tests

Test

Descriptive Statistics

Test

One Sample Mean z-Test (Optional)

Descriptive Statistics

Test

8.3 Paired Means

8.3.1 Confidence Intervals

Minitab: Paired Means Confidence Interval

8.3.2 Hypothesis Testing

Minitab: Paired Means Test

Paired t: SATM, SATV

Descriptive Statistics

Estimation for Paired Difference

Test

Try It! Marriage Age (Summarized Data)

Estimation for Paired Difference

Test

8.4 Summary